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texas instrument paperTRANSCRIPT
Texas Instruments Placement Paper (Reasoning)
1. If the date is written as MMDDYYYY, and then 10022001, ie)oct 2 2001 is a
palindrome. which is the immediate palindrome before that date.
2. There r 2 trains travelling at 100 MPH each. there is a 200Mile tunnel. the
trains travel in opposite directions. when the trains enter, a supersonic bee starts
at the tunnel, touches the other train, comes back touches the first train, turns
again and touches the other train and so on until the trains collide. Find the
distance travelled by the bee.
3. It is 12.00 o clock. Find how often the minute hand meets the hour hand.
4. Find the maximum product using positive integers the sum of which is 100. Dont
know if i am sure with this one. sorry. i didnt do this.
5. In a train Art is found dead. hours later 4 people are queried,
Blonde says i am innocent, i didnt speak with Art
White says i am innocent, Blonde spoke with Art
Old says i am innocent, the brunette killed Art
Brunette says i am innocent, one of the men killed Art
The inspector says simple, 4 true statements, 4 false statements.
I know the killer, very simple!!!
Find out the killer. (only one is involved.)
6. 2 maths geniuses meet after 20 yrs.one says i am married. and have 3
daughters. the product of their ages is 72. the sum is the same as the house no.
other says, but I cant find it out. The other says. . simple. ok, my eldest daughter
has just started the piano classes.Find their ages.This one is there in previous
question papers also.
7. 5 sea pirates have 100 gold coins and want to share it. They propose a plan. The
senior most one has to propose an idea, if at least 50 % agree, the coins r shared
accordingly. else the senior is killed and the next senior most is asked to present
a plan and so on .note, all the guys r very clever and very greedy and dont want
to lose the coins, and dont want to die. Form a way to share the money.
I didnt do this....no time
ANSWERS
1. I worked it out like this.. before 2001, 2000 cannot be taken as MMDD becomes
0002.also anything after 14 century cant be considered as 1401 would mean
1041, there r only a max of 31 days a month. so the century nos must be only
13(reverse for 31)so the yr is 13xx. now 1399to 1391 is not possible as reverse
means 99 to 19 but max no of months is 12. so 1390 becomes 09311390... but
9th month has only 30 days(lucky i remembered it).. so 1380 is the most recent
palindrome ... which beocomes 08311380.SOln is 08 31 1380.
2. This one i really screwed up royally. it took me 25 minutes to solve.. i took the
average speeds, the distance travelled by the trains and the bee, took ratio of
speeds and distance, subtracting it from the tunnels length for every time the bee
touched the trains. phew lot of calculation.. and got the ans as 1000.But when i
rechecked it a thought struck me... so silly of me. The trains collide at the middle.
ie)after 100 miles. that means after 1 hour. so the bee travels for 1 hour touching
each train till they collide. its speed being 1000 mph, it travelled a 1000 miles..
3. At 12.00 the minute and hour hand are together. but at 1.00 they are 5 mins
apart. so in 60 minutes the gain is 55 mins for the min hand. to be together it
has to gain another 5 more mins. so the time taken to gain 5 mins is
60*5/55==5.454 So the time taken to meet again is 65.454 So the hands meet
every 65.454 minutes
1. nt fact(int n) int fact_ii(int n) { { if (n >0) int fact_val = 1, i = 0; return n*fact (n-1); for (i = 1; i <= n ; i++) else fact_val = fact_val*i; return 1; return fact_val; } } (a) (i) only (b) (ii) only (c) both (d) None of these
2. (2) Which of the following is a semantic error? (a) Division by zero (b) Missing of a semicolon at the end of a statement (c) Assigning a single precision real value to a long integer (d) All the above
3. (3) If the following code segment is to count the no of zero’s in the given integer ‘x’ in its binary representation, what is to replaced by CONDITION? int i, count, x; for(i=0, count=0;i<16;i++) if(CONDITION) count++; a. (x & (1 << i)) b. (x & (1 << i))
c. (x && (1 << i)) d. None of the above.
4. A graph was given and asked to find out the minimum no of colors required to color the graph? (a) 1 (b) 2 (c) 3 (d) The graph can’t be colored
5. Inorder traversal of a binary tree is d c e b a and preorder traversal of a binary tree is b c d e a. Find out the post order traversal? (6) Which is not an advantage of using subroutines? (a) Easier maintenance (b) Runtime reduces (c) Storage space reduces (d) Modularity
6. A k-diagonal matrix is a n *n square matrix in which the elements on the principal diagonal and k diagonals above the principal diagonal and k diagonals below the principal diagonal only have none zero elements. Other elements are zero’s. In order to save the space, the non zero elements are stored in a one dimensional array. The no of locations in this array are: (a) n*(n-k-1)/2 (b) n*(n-1) – (n-k)(n-k-2) (c) n*(n-1) – (n-k)(n-k-2) (d) n*n – (n-k-1)
7. A 1D - array A whose size is N is given and is divided into P partitions and an element x is to be searched in the array. Each partition is given to oneprocessor. The elements are searched with in the partition using a binary search. What is the time complexity of the algorithm? (a) O (N/P) (b) O (P) (c) O ( log (N/P)) (d) O ( log (N/P)) + O (P log P)
8. Write the code segment to insert an element p into the linked list after an element q? Make necessary pointer adjustments?
Ans: p.next = q.next; q.next = p;
9. Given a matrix A. What is the minimum no of multiplications do u need to compute A10 ? (a) 9 (b) 5 (c) 4 (d) 6
1. if a 5-stage pipe-line is flushed and then we have to execute 5 and 12
instructions respectively then no. of cycles will be
a. 5 and 12
b. 6 and 13
c. 9 and 16
d.none
2. k-map
ab
----------
c 1 x 0 0
1 x 0 x
solve it
a. A.B
B. ~A
C. ~B
D. A+B
3. CHAR A[10][15] AND INT B[10][15] IS DEFINED WHAT'S THE ADDRESS OF
A[3][4] AND B[3][4] IF ADDRESS OD A IS OX1000 AND B IS 0X2000
A. 0X1030 AND 0X20C3
B. OX1031 AND OX20C4
AND SOME OTHERS..
4. int f(int *a)
{
int b=5;
a=&b;
}
main()
{
int i;
printf("\n %d",i);
f(&i);
printf("\n %d",i);
}
what's the output .
1.10,5
2,10,10
c.5,5
d. none
5. main()
{
int i;
fork();
fork();
fork();
printf("----");
}
how many times the printf will be executed .
a.3
b. 6
c.5
d. 8
6. void f(int i)
{
int j;
for (j=0;j<16;j++)
{
if (i & (0x8000>>j))
printf("1");
else
printf("0");
}
}
what's the purpose of the program
a. its output is hex representation of i
b. bcd
c. binary
d. decimal
7. #define f(a,b) a+b
#define g(a,b) a*b
main()
{
int m;
m=2*f(3,g(4,5));
printf("\n m is %d",m);
}
what's the value of m
a.70
b.50
c.26
d. 69
8. main()
{
char a[10];
strcpy(a,"\0");
if (a==NULL)
printf("\a is null");
else
printf("\n a is not null");}
what happens with it .
a. compile time error.
b. run-time error.
c. a is null
d. a is not null.
9. char a[5]="hello"
a. in array we can't do the operation .
b. size of a is too large
c. size of a is too small
d. nothing wrong with it .
10. local variables can be store by compiler
a. in register or heap
b. in register or stack
c .in stack or heap .
d. global memory.
11. average and worst time complexity in a sorted binary tree is
12. a tree is given and ask to find its meaning (parse-tree)
(expression tree)
ans. ((a+b)-(c*d)) ( not confirmed)
13. convert 40.xxxx into binary .
14. global variable conflicts due to multiple file occurance
is resolved during
a. compile-time
b. run-time
c. link-time
d. load-time
15. two program is given of factorial. one with recursion and one without
recursion .question was which program won't run for very big no. input because
of stack overfow .
a. i only (ans.)
b. ii only
c. i& ii both .
c. none
16. struct a
{
int a;
char b;
int c;
}
union b
{
char a;
int b;
int c;
};
which is correct .
a. size of a is always diff. form size of b.(ans.)
b. size of a is always same form size of b.
c. we can't say anything because of not-homogeneous (not in ordered)
d. size of a can be same if ...
1. (steeper transition) by:
a. Increasing W/L of PMOS transistor
b. Increasing W/L of NMOS transistor
c. Increasing W/L of both transistors by the same factor
d. Decreasing W/L of both transistor by the same factor
2. Minimum number of 2-input NAND gates that will be required to
implement the function: Y = AB + CD + EF is
a. 4
b. 5
c. 6
d. 7
3. Consider a two-level memory hierarchy system M1 & M2. M1 is accessed
first and on miss M2 is accessed. The access of M1 is 2 nanoseconds and the miss
penalty (the time to get the data from M2 in case of a miss) is 100 nanoseconds.
The probability that a valid data is found in M1 is 0.97. The average memory
access time is:
a. 4.94 nanoseconds
b. 3.06 nanoseconds
c. 5.00 nanoseconds
d. 5.06 nanoseconds
4. Interrupt latency is the time elapsed between:
a. Occurrence of an interrupt and its detection by the CPU
b. Assertion of an interrupt and the start of the associated ISR
c. Assertion of an interrupt and the completion of the associated ISR
d. Start and completion of associated ISR
5. Which of the following is true for the function (A.B + A’.C + B.C)
a. This function can glitch and can be further reduced
b. This function can neither glitch nor can be further reduced
c. This function can glitch and cannot be further reduced
d. This function cannot glitch but can be further reduced
6. For the two flip-flop configuration below, what is the relationship of the
output at B to the clock frequency?
a. Output frequency is 1/4th the clock frequency, with 50% duty cycle
b. Output frequency is 1/3rd the clock frequency, with 50% duty cycle
c. Output frequency is 1/4th the clock frequency, with 25% duty cycle
d. Output frequency is equal to the clock frequency
7. The voltage on Node B is: a. 0 b. 10 c. –10
8.A CPU supports 250 instructions. Each instruction op-code has these fields:
a. The instruction type (one among 250) b. A conditional register specification c. 3 register operands d. Addressing mode specification for both source operands
9. The CPU has 16 registers and supports 5 addressing modes. What is the instruction
op-code length in bits?
32
24
30
36
10. In the iterative network shown, the output Yn of any stage N is 1 if the total number of 1s at the inputs starting from the first stage to the Nth stage is odd. (Each identical box in the iterative network has two inputs and two outputs). The optimal logic structure for the box consists of:
a. One AND gate and one NOR gate b. One NOR gate and one NAND gate c. Two XNOR gates d. One XOR gate
11.Consider a circuit with N logic nets. If each net can be stuck-at either values 0 and 1, in how many ways can the circuit be faulty such that only one net in it can be faulty, and such that up-to all nets in it can be faulty?
a. 2 and 2N b. N and 2^N c. 2N and 3^N-1 d. 2N and 3N
12. In the circuit shown, all the flip-flops are identical. If the set-up time is 2 ns, clock->Q delay is 3 ns and hold time is 1 ns, what is the maximum frequency of operation for the circuit? a. 200 MHz b. 333 MHz c. 250 MHz d. None of the above
13. Which of the following statements is/are true?
I. Combinational circuits may have feedback, sequential circuits do not.
II. Combinational circuits have a ‘memory-less’ property, sequential circuits do not.
III. Both combinational and sequential circuits must be controlled by an external clock.
a. I only b. II and III only c. I and II only
d. II only
14.Consider an alternate binary number representation scheme, wherein the number of ones M, in a word of N bits, is always the same. This scheme is called the M-out-of-N coding scheme. If M=N/2, and N=8, what is the efficiency of this coding scheme as against the regular binary number representation scheme? (As a hint, consider that the number of unique words represent able in the latter representation with N bits is 2^N. Hence the efficiency is 100%)
a. Close to 30% b. Close to 50% c. Close to 70% d. Close to 100%
15.A CPU supports 4 interrupts- I1, I2, I3 and I4. It supports priority of interrupts. Nested interrupts are allowed if later interrupt is higher priority than previous one. During a certain period of time, we observe the following sequence of entry into and exit from the interrupt service routine:
a. I1-start---I2-start---I2-end---I4-start---I3-start---I3-end---I4-end---I1-end
From this sequence, what can we infer about the interrupt routines?
a. I3 > I4 > I2 > I1 b. I4 > I3 > I2 > I1 c. I2 > I1; I3 > I4 > I1 d. I2 > I1, I3 > I4 > I2 > I1
16. I decide to build myself a small electric kettle to boil my cup of tea. I need 200 ml of water for my cup of tea. Assuming that typical tap water temperature is 25 C and I want the water boiling in exactly one minute, then what is the wattage required for the heating element?
[Assume: Boiling point of water is 100 C, 1 Calorie (heat required to change 1 gm of water by 1 C)= 4 joules, 1 ml of water weighs 1 gm.]
a. Data given is insufficient b. 800 W c. 300 W d. 1000 W e. 250 W
17.The athletics team from REC Trichy is traveling by train. The train slows down, (but does not halt) at a small wayside station that has a 100 mts long platform. The sprinter (who can run 100 mts in 10 sec) decides to jump down and get a newspaper and some idlis. He jumps out just as his compartment enters the platform and spends 5 secs buying his newspaper that is at the point where he jumped out. He then sprints along the platform to buy idlis that is another 50 mts. He spends another 5 secs buying the idlis. He is now just 50 mts from the other end of the platform where the train is moving out. He begins running in the direction of the train and the only other open door in his train is located 50 mts behind the door from where he jumped. At what(uniform) speed should the train be traveled if he just misses jumping into the open door at the very edge of the platform?
a. Make the following assumptions He always runs at his peak speed uniformly
The train travels at uniform speed He does not wait (other than for the idlis & newspaper) or run
baclwards b. Data given is insufficient
c. 4 m/s d. 5 m/s e. 7.5 m/s f. 10 m/s
18.State which of the following gate combinations does not form a universal logic set:
a. 2-input AND + 2-input OR b. 2-to-1 multiplexer c. 2-input XOR + inverter d. 3-input NAND
19.For the circuit shown below, what should the function F be, so that it produces an output of the same frequency (function F1), and an output of double the frequency (function F2).
a. F1= NOR gate and F2= OR gate b. F1=NAND gate and F2= AND gate
c. F1=AND gate and F2=XOR gate d. None of the above
20.The FSM (finite state machine) below starts in state Sa, which is the reset state, and detects a particular sequence of inputs leading it to state Sc. FSMs have a few characteristics. An autonomous FSM has no inputs. For a Moore FSM, the output depends on the present state alone. For a Mealy FSM, the output depends on the present state as well as the inputs. Which of the statements best describes the FSM below?
a. It has two states and is autonomous b. The information available is insufficient
c. It is a Mealy machine with three states d. It is a Moor machine with three states
21In the circuit given below, the switch is opened at time t=0. Voltage across the capacitor at t=infinity is:
a. 2V b. 3V c. 5V d. 7V
22.What is the functionality represented by the following circuit?
a. y= ! (b+ac) b. y= ! (a+bc) c. y= ! (a(b+c)) d. y= ! (a+b+c)
23.The value (0xdeadbeef) needs to stored at address 0x400. Which of the below ways will the memory look like in a big endian machine: 0x403 0x402 0x401 0x400
a. be ef de ad
b. ef be ad de
c. fe eb da ed
d. ed da eb fe
24. In a given CPU-memory sub-system, all accesses to the memory take two cycles. Accesses to memories in two consecutive cycles can therefore result in incorrect data transfer. Which of the following access mechanisms guarantees correct data transfer?
a. A read operation followed by a write operation in the next cycle. b. A write operation followed by a read operation in the next cycle.
c. A NOP between every successive reads & writes d. None of the above
25.An architecture saves 4 control registers automatically on function entry (and restores them on function return). Save of each registers costs 1 cycle (so does restore). How many cycles are spent in these tasks (save and restore) while running the following un-optimized code with n=5:
a. Void fib(int n)
{
if((n==0) || (n==1)) return 1;
return(fib(n-1) + fib(n-2));
}
a. 120 b. 80 c. 125 d. 128
26.The maximum number of unique Boolean functions F(A,B), realizable for a two input (A,B) and single output (Z) circuit is:
a. 2 b. 6 c. 8
Can we declare a static function as virtual?
Ans: No. The virtual function mechanism is used on the specific object that
determines which virtual function to call. Since the static functions are not any
way related to objects, they cannot be declared as virtual.
2. Can user-defined object be declared as static data member of another class?
Ans: Yes. The following code shows how to initialize a user-defined object.
#include
class test
{
int i ;
public :
test ( int ii = 0 )
{
i = ii ;
}
} ;
class sample
{
static test s ;
} ;
test sample::s ( 26 ) ;
Here we have initialized the object s by calling the one-argument constructor.
We can use the same convention to initialize the object by calling multiple-
argument constructor.
3. What is forward referencing and when should it be used?
Ans: Consider the following program:
class test
{
public :
friend void fun ( sample, test ) ;
} ;
class sample
{
public :
friend void fun ( sample, test ) ;
} ;
void fun ( sample s, test t )
{
// code
}
void main( )
{
sample s ;
test t ;
fun ( s, t ) ;
}
This program would not compile. It gives an error that sample is undeclared
identifier in the statement friend void fun ( sample, test ) ; of the class test. This
is so because the class sample is defined below the class test and we are using it
before its definition. To overcome this error we need to give forward reference of
the class sample before the definition of class test. The following statement is
the forward reference of class sample. Forward referencing is generally required
when we make a class or a function as a friend.
4. The istream_withassign class has been derived from the istream class and
overloaded assignment operator has been added to it. The _withassign classes
are much like their base classes except that they include overloaded assignment
operators. Using these operators the objects of the _withassign classes can be
copied. The istream, ostream, and iostream classes are made uncopyable by
making their overloaded copy constructor and assignment operators private.
5. How do I write my own zero-argument manipulator that should work same as
hex?
Ans: This is shown in following program.
#include
ostream& myhex ( ostream &o )
{
o.setf ( ios::hex) ;
return o ;
}
void main( )
{
cout << endl << myhex << 2000 ;
}
6.We all know that a const variable needs to be initialized at the time of
declaration. Then how come the program given below runs properly even when
we have not initialized p?
#include
void main( )
{
const char *p ;
p = "A const pointer" ;
cout << p ;
}
Ans: The output of the above program is 'A const pointer'. This is because in this
program p is declared as 'const char*' which means that value stored at p will be
constant and not p and so the program works properly
7. How do I refer to a name of class or function that is defined within a
namespace?
Ans: There are two ways in which we can refer to a name of class or function
that is defined within a namespace: Using scope resolution operator through the
using keyword. This is shown in following example:
namespace name1
{
class sample1
{
// code
} ;
}
namespace name2
{
class sample2
{
// code
} ;
}
using namespace name2 ;
void main( )
{
name1::sample1 s1 ;
sample2 s2 ;
}
Here, class sample1 is referred using the scope resolution operator. On the other
hand we can directly refer to class sample2 because of the statement using
namespace name2 ; the using keyword declares all the names in the namespace
to be in the current scope. So we can use the names without any qualifiers.
8. While overloading a binary operator can we provide default values?
Ans: No!. This is because even if we provide the default arguments to the
parameters of the overloaded operator function we would end up using the
binary operator incorrectly. This is explained in the following example:
sample operator + ( sample a, sample b = sample (2, 3.5f ) )
{
}
void main( )
{
sample s1, s2, s3 ;
s3 = s1 + ; // error
}
9. How do I carry out conversion of one object of user-defined type to another?
Ans: To perform conversion from one user-defined type to another we need to
provide conversion function. Following program demonstrates how to provide
such conversion function.
class circle
{
private :
int radius ;
public:
circle ( int r = 0 )
{
radius = r ;
}
} ;
class rectangle
{
private :
int length, breadth ;
public :
rectangle( int l, int b )
{
length = l ;
breadth = b ;
}
operator circle( )
{
return circle ( length ) ;
}
} ;
void main( )
{
rectangle r ( 20, 10 ) ;
circle c;
c = r ;
}
Here, when the statement c = r ; is executed the compiler searches for an
overloaded assignment operator in the class circle which accepts the object of
type rectangle. Since there is no such overloaded assignment operator, the
conversion operator function that converts the rectangle object to the circle
object is searched in the rectangle class. We have provided such a conversion
function in the rectangle class. This conversion operator function returns a circle
object. By default conversion operators have the name and return type same as
the object type to which it converts to. Here the type of the object is circle and
hence the name of the operator function as well as the return type is circle.
10. How do I write code that allows to create only one instance of a class?
Ans: This is shown in following code snippet.
#include
class sample
{
static sample *ptr ;
private:
sample( )
{
}
public:
static sample* create( )
{
if ( ptr == NULL )
ptr = new sample ;
return ptr ;
}
} ;
sample *sample::ptr = NULL ;
void main( )
{
sample *a = sample::create( ) ;
sample *b = sample::create( ) ;
}
Here, the class sample contains a static data member ptr, which is a pointer
to the object of same class. The constructor is private which avoids us from
creating objects outside the class. A static member function called create( ) is
used to create an object of the class. In this function the condition is checked
whether or not ptr is NULL, if it is then an object is created dynamically and its
address collected in ptr is returned. If ptr is not NULL, then the same address is
returned. Thus, in main( ) on execution of the first statement one object of
sample gets created whereas on execution of second statement, b holds the
address of the first object. Thus, whatever number of times you call create( )
function, only one object of sample class will be available.
11. How do I write code to add functions, which would work as get and put
properties of a class?
Ans: This is shown in following code.
#include
class sample
{
int data ;
public:
__declspec ( property ( put = fun1, get = fun2 ) )
int x ;
void fun1 ( int i )
{
if ( i < 0 )
data = 0 ;
else
data = i ;
}
int fun2( )
{
return data ;
}
} ;
void main( )
{
sample a ;
a.x = -99 ;
cout << a.x ;
}
Here, the function fun1( ) of class sample is used to set the given integer value
into data, whereas fun2( ) returns the current value of data. To set these
functions as properties of a class we have given the statement as shown below:
__declspec ( property ( put = fun1, get = fun2 )) int x ;
As a result, the statement a.x = -99 ; would cause fun1( ) to get called to set
the value in data. On the other hand, the last statement would cause fun2( ) to
get called to return the value of data.
12. How do I write code to make an object work like a 2-D array?
Ans: Take a look at the following program.
#include
class emp
{
public :
int a[3][3] ;
emp( )
{
int c = 1 ;
for ( int i = 0 ; i <= 2 ; i++ )
{
for ( int j = 0 ; j <= 2 ; j++ )
{
a[i][j] = c ;
c++ ;
}
}
}
int* operator[] ( int i )
{
return a[i] ;
}
} ;
void main( )
{
emp e ;
cout << e[0][1] ;
}
The class emp has an overloaded operator [ ] function. It takes one argument an
integer representing an array index and returns an int pointer. The statement
cout << e[0][1] ; would get converted into a call to the overloaded [ ] function
as e.operator[ ] ( 0 ). 0 would get collected in i. The function would return a[i]
that represents the base address of the zeroeth row. Next the statement would
get expanded as base address of zeroeth row[1] that can be further expanded
as *( base address + 1 ). This gives us a value in zeroth row and first column.
13. What are formatting flags in ios class?
Ans: The ios class contains formatting flags that help users to format the stream
data. Formatting flags are a set of enum definitions. There are two types of
formatting flags:
On/Off flags
Flags that work in-group
The On/Off flags are turned on using the setf( ) function and are turned off using
the unsetf( ) function. To set the On/Off flags, the one argument setf( ) function
is used. The flags working in groups are set through the two-argument setf( )
function. For example, to left justify a string we can set the flag as,
cout.setf ( ios::left ) ;
cout << "KICIT Nagpur" ;
To remove the left justification for subsequent output we can say,
cout.unsetf ( ios::left ) ;
The flags that can be set/unset include skipws, showbase, showpoint,
uppercase, showpos, unitbuf and stdio. The flags that work in a group can have
only one of these flags set at a time.
14. What is the purpose of ios::basefield in the following statement?
cout.setf ( ios::hex, ios::basefield ) ;
Ans: This is an example of formatting flags that work in a group. There is a flag
for each numbering system (base) like decimal, octal and hexadecimal.
Collectively, these flags are referred to as basefield and are specified by
ios::basefield flag. We can have only one of these flags on at a time. If we set
the hex flag as setf ( ios::hex ) then we will set the hex bit but we won't clear
the dec bit resulting in undefined behavior. The solution is to call setf( ) as setf (
ios::hex, ios::basefield ). This call first clears all the bits and then sets the hex
bit.
15. Can we get the value of ios format flags?
Ans: Yes! The ios::flags( ) member function gives the value format flags. This
function takes no arguments and returns a long ( typedefed to fmtflags) that
contains the current format flags.
16. Is there any function that can skip certain number of characters present in
the input stream?
Ans: Yes! This can be done using cin::ignore( ) function. The prototype of this
function is as shown below:
istream& ignore ( int n = 1, int d =EOF ) ;
Sometimes it happens that some extra characters are left in the input stream
while taking the input such as, the ?\n? (Enter) character. This extra character is
then passed to the next input and may pose problem.
To get rid of such extra characters the cin::ignore( ) function is used. This is
equivalent to fflush ( stdin ) used in C language. This function ignores the first n
characters (if present) in the input stream, stops if delimiter d is encountered.
17. Write a program that implements a date class containing day, month and
year as data members. Implement assignment operator and copy constructor in
this class.
Ans: This is shown in following program:
#include
class date
{
private :
int day ;
int month ;
int year ;
public :
date ( int d = 0, int m = 0, int y = 0 )
{
day = d ;
month = m ;
year = y ;
}
// copy constructor
date ( date &d )
{
day = d.day ;
month = d.month ;
year = d.year ;
}
// an overloaded assignment operator
date operator = ( date d )
{
day = d.day ;
month = d.month ;
year = d.year ;
return d ;
}
void display( )
{
cout << day << "/" << month << "/" << year ;
}
} ;
void main( )
{
date d1 ( 25, 9, 1979 ) ;
date d2 = d1 ;
date d3 ;
d3 = d2 ;
d3.display( ) ;
}
18. When should I use unitbuf flag?
Ans: The unit buffering (unitbuf) flag should be turned on when we want to
ensure that each character is output as soon as it is inserted into an output
stream. The same can be done using unbuffered output but unit buffering
provides a better performance than the unbuffered output.
19.What are manipulators?
Ans: Manipulators are the instructions to the output stream to modify the output
in various ways. The manipulators provide a clean and easy way for formatted
output in comparison to the formatting flags of the ios class. When manipulators
are used, the formatting instructions are inserted directly into the stream.
Manipulators are of two types, those that take an argument and those that
don?t.
20. What is the difference between the manipulator and setf( ) function?
Ans: The difference between the manipulator and setf( ) function are as follows:
The setf( ) function is used to set the flags of the ios but manipulators directly
insert the formatting instructions into the stream. We can create user-defined
manipulators but setf( ) function uses data members of ios class only. The flags
put on through the setf( ) function can be put off through unsetf( ) function.
Such flexibility is not available with manipulators.
21. How do I get the current position of the file pointer?
Ans: We can get the current position of the file pointer by using the tellp( )
member function of ostream class or tellg( ) member function of istream class.
These functions return (in bytes) positions of put pointer and get pointer
respectively.
22. What are put and get pointers?
Ans: These are the long integers associated with the streams. The value present
in the put pointer specifies the byte number in the file from where next write
would take place in the file. The get pointer specifies the byte number in the file
from where the next reading should take place.
23. What do the nocreate and noreplace flag ensure when they are used for
opening a file?
Ans: nocreate and noreplace are file-opening modes. A bit in the ios class
defines these modes. The flag nocreate ensures that the file must exist before
opening it. On the other hand the flag noreplace ensures that while opening a
file for output it does not get overwritten with new one unless ate or app is set.
When the app flag is set then whatever we write gets appended to the existing
file. When ate flag is set we can start reading or writing at the end of existing
file.
24. What is the limitation of cin while taking input for character array?
Ans: To understand this consider following statements,
char str[5] ;
cin >> str ;
While entering the value for str if we enter more than 5 characters then there is
no provision in cin to check the array bounds. If the array overflows, it may be
dangerous. This can be avoided by using get( ) function. For example, consider
following statement,
cin.get ( str, 5 ) ;
On executing this statement if we enter more than 5 characters, then get( )
takes only first five characters and ignores rest of the characters. Some more
variations of get( ) are available, such as shown below:
get ( ch ) ? Extracts one character only
get ( str, n ) ? Extracts up to n characters into str
get ( str, DELIM ) ? Extracts characters into array str until specified delimiter
(such as '\n'). Leaves delimiting character in stream.
get ( str, n, DELIM ) ? Extracts characters into array str until n characters or
DELIM character, leaving delimiting character in stream.
25. What is the purpose of istream class?
Ans: The istream class performs activities specific to input. It is derived from the
ios class. The most commonly used member function of this class is the
overloaded >> operator which can extract values of all basic types. We can
extract even a string using this operator.
26. Would the following code work?
#include
void main( )
{
ostream o ;
o << "Dream. Then make it happen!" ;
}
Ans: No! This is because we cannot create an object of the ostream class since
its constructor and copy constructor are declared private.
27. Can we use this pointer inside static member function?
Ans: No! The this pointer cannot be used inside a static member function. This is
because a static member function is never called through an object.
28. What is strstream?
Ans: strstream is a type of input/output stream that works with the memory. It
allows using section of the memory as a stream object. These streams provide
the classes that can be used for storing the stream of bytes into memory. For
example, we can store integers, floats and strings as a stream of bytes. There
are several classes that implement this in-memory formatting. The class
ostrstream derived from ostream is used when output is to be sent to memory,
the class istrstream derived from istream is used when input is taken from
memory and strstream class derived from iostream is used for memory objects
that do both input and output. Ans: When we want to retrieve the streams of
bytes from memory we can use istrestream. The following example shows the
use of istrstream class.
#include
void main( )
{
int age ;
float salary ;
char name[50] ;
char str[] = "22 12004.50 K. Vishwanatth" ;
istrstream s ( str ) ;
s >> age >> salary >> name ;
cout << age << endl << salary << endl << name ;
cout << endl << s.rdbuf( ) ;
}
Here, s is the object of the class istrstream. When we are creating the object s,
the constructor of istrstream gets called that receives a pointer to the zero
terminated character array str. The statement s >> age >> salary >> name ;
extracts the age, salary and the name from the istrstream object s. However,
while extracting the name, only the first word of name gets extracted. The
balance is extracted using rdbuf( ).
29. When the constructor of a base class calls a virtual function, why doesn't the
override function of the derived class gets called?
Ans: While building an object of a derived class first the constructor of the base
class and then the constructor of the derived class gets called. The object is said
an immature object at the stage when the constructor of base class is called.
This object will be called a matured object after the execution of the constructor
of the derived class. Thus, if we call a virtual function when an object is still
immature, obviously, the virtual function of the base class would get called. This
is illustrated in the following example.
#include
class base
{
protected :
int i ;
public :
base ( int ii = 0 )
{
i = ii ;
show( ) ;
}
virtual void show( )
{
cout << "base's show( )" << endl ;
}
} ;
class derived : public base
{
private :
int j ;
public :
derived ( int ii, int jj = 0 ) : base ( ii )
{
j = jj ;
show( ) ;
}
void show( )
{
cout << "derived's show( )" << endl ;
}
} ;
void main( )
{
derived dobj ( 20, 5 ) ;
}
The output of this program would be:
base's show( )
derived's show( )
30. Can I have a reference as a data member of a class? If yes, then how do I
initialise it?
Ans: Yes, we can have a reference as a data member of a class. A reference as a
data member of a class is initialized in the initialization list of the constructor.
This is shown in following program.
#include
class sample
{
private :
int& i ;
public :
sample ( int& ii ) : i ( ii )
{
}
void show( )
{
cout << i << endl ;
}
} ;
void main( )
{
int j = 10 ;
sample s ( j ) ;
s.show( ) ;
}
Here, i refers to a variable j allocated on the stack. A point to note here is that
we cannot bind a reference to an object passed to the constructor as a value. If
we do so, then the reference i would refer to the function parameter (i.e.
parameter ii in the constructor), which would disappear as soon as the function
returns, thereby creating a situation of dangling reference.
31. Why does the following code fail?
#include
class sample
{
private :
char *str ;
public :
sample ( char *s )
{
strcpy ( str, s ) ;
}
~sample( )
{
delete str ;
}
} ;
void main( )
{
sample s1 ( "abc" ) ;
}
Ans: Here, through the destructor we are trying to deal locate memory, which
has been allocated statically. To remove an exception, add following statement
to the constructor.
sample ( char *s )
{
str = new char[strlen(s) + 1] ;
strcpy ( str, s ) ;
}
Here, first we have allocated memory of required size, which then would get deal
located through the destructor.
32. assert( ) macro...
We can use a macro called assert( ) to test for conditions that should not occur
in a code. This macro expands to an if statement. If test evaluates to 0, assert
prints an error message and calls abort to abort the program.
#include
#include
void main( )
{
int i ;
cout << "\nEnter an integer: " ;
cin >> i ;
assert ( i >= 0 ) ;
cout << i << endl ;
}
33. Why it is unsafe to deal locate the memory using free( ) if it has been
allocated using new?
Ans: This can be explained with the following example:
#include
class sample
{
int *p ;
public :
sample( )
{
p = new int ;
}
~sample( )
{
delete p ;
}
} ;
void main( )
{
sample *s1 = new sample ;
free ( s1 ) ;
sample *s2 = ( sample * ) malloc ( sizeof ( sample ) ) ;
delete s2 ;
}
The new operator allocates memory and calls the constructor. In the constructor
we have allocated memory on heap, which is pointed to by p. If we release the
object using the free( ) function the object would die but the memory allocated
in the constructor would leak. This is because free( ) being a C library function
does not call the destructor where we have deal located the memory.
As against this, if we allocate memory by calling malloc( ) the constructor would
not get called. Hence p holds a garbage address. Now if the memory is deal
located using delete, the destructor would get called where we have tried to
release the memory pointed to by p. Since p contains garbage this may result in
a runtime error.
34. Can we distribute function templates and class templates in object libraries?
Ans: No! We can compile a function template or a class template into object
code (.obj file). The code that contains a call to the function template or the
code that creates an object from a class template can get compiled. This is
because the compiler merely checks whether the call matches the declaration (in
case of function template) and whether the object definition matches class
declaration (in case of class template). Since the function template and the class
template definitions are not found, the compiler leaves it to the linker to restore
this. However, during linking, linker doesn't find the matching definitions for the
function call or a matching definition for object creation. In short the expanded
versions of templates are not found in the object library. Hence the linker
reports error.
35. What is the difference between an inspector and a mutator ?
Ans: An inspector is a member function that returns information about an
object's state (information stored in object's data members) without changing
the object's state. A mutator is a member function that changes the state of an
object. In the class Stack given below we have defined a mutator and an
inspector.
class Stack
{
public :
int pop( ) ;
int getcount( ) ;
}
In the above example, the function pop( ) removes top element of stack thereby
changing the state of an object. So, the function pop( ) is a mutator. The
function getcount( ) is an inspector because it simply counts the number of
elements in the stack without changing the stack.
36. Namespaces:
The C++ language provides a single global namespace. This can cause problems
with global name clashes. For instance, consider these two C++ header files:
// file1.h
float f ( float, int ) ;
class sample { ... } ;
// file2.h
class sample { ... } ;
With these definitions, it is impossible to use both header files in a single
program; the sample classes will clash. A namespace is a declarative region that
attaches an additional identifier to any names declared inside it. The additional
identifier thus avoids the possibility that a name will conflict with names declared
elsewhere in the program. It is possible to use the same name in separate
namespaces without conflict even if the names appear in the same translation
unit. As long as they appear in separate namespaces, each name will be unique
because of the addition of the namespace identifier. For example:
// file1.h
namespace file1
{
float f ( float, int ) ;
class sample { ... } ;
}
// file2.h
namespace file2
{
class sample { ... } ;
}
Now the class names will not clash because they become file1::sample and
file2::sample, respectively.
37. What would be the output of the following program?
#include
class user
{
int i ;
float f ;
char c ;
public :
void displaydata( )
{
cout << endl << i << endl << f << endl << c ;
}
} ;
void main( )
{
cout << sizeof ( user ) ;
user u1 ;
cout << endl << sizeof ( u1 ) ;
u1.displaydata( ) ;
}
Ans: The output of this program would be,
9 or 7
9 or 7
Garbage
Garbage
Garbage
Since the user class contains three elements, int, float and char its size would be
9 bytes (int-4, float-4, char-1) under Windows and 7 bytes (int-2, float-4, char-
1) under DOS. Second output is again the same because u1 is an object of the
class user. Finally three garbage values are printed out because i, f and c are not
initialized anywhere in the program.
Note that if you run this program you may not get the answer shown here. This
is because packing is done for an object in memory to increase the access
efficiency. For example, under DOS, the object would be aligned on a 2-byte
boundary. As a result, the size of the object would be reported as 6 bytes. Unlike
this, Windows being a 32-bit OS the object would be aligned on a 4-byte
boundary. Hence the size of the object would be reported as 12 bytes. To force
the alignment on a 1-byte boundary, write the following statement before the
class declaration.
#pragma pack ( 1 )
38. Write a program that will convert an integer pointer to an integer and vice-
versa.
Ans: The following program demonstrates this.
#include
void main( )
{
int i = 65000 ;
int *iptr = reinterpret_cast ( i ) ;
cout << endl << iptr ;
iptr++ ;
cout << endl << iptr ;
i = reinterpret_cast ( iptr ) ;
cout << endl << i ;
i++ ;
cout << endl << i ;
}
39. What is a const_cast?
Ans. The const_cast is used to convert a const to a non-const. This is shown in
the following
program:
#include
void main( )
{
const int a = 0 ;
int *ptr = ( int * ) &a ; //one way
ptr = const_cast_ ( &a ) ; //better way
}
Here, the address of the const variable a is assigned to the pointer to a non-
const variable. The const_cast is also used when we want to change the data
members of a class inside the const member functions. The following code
snippet shows this:
class sample
{
private:
int data;
public:
void func( ) const
{
(const_cast (this))->data = 70 ;
}
} ;
40. What is forward referencing and when should it be used?
Ans: Forward referencing is generally required when we make a class or a
function as a friend.
Consider following program:
class test
{
public:
friend void fun ( sample, test ) ;
} ;
class sample
{
public:
friend void fun ( sample, test ) ;
} ;
void fun ( sample s, test t )
{
// code
}
void main( )
{
sample s ;
test t ;
fun ( s, t ) ;
}
On compiling this program it gives error on the following statement of test class.
It gives an error that sample is undeclared identifier. friend void fun ( sample,
test );
This is so because the class sample is defined below the class test and we are
using it before its definition. To overcome this error we need to give forward
reference of the class sample before the definition of class test. The following
statement is the forward reference of class sample.
class sample ;
41. How would you give an alternate name to a namespace?
Ans: An alternate name given to namespace is called a namespace-alias.
namespace-alias is generally used to save the typing effort when the names of
namespaces are very long or complex. The following syntax is used to give an
alias to a namespace.
namespace myname = my_old_very_long_name ;
42. Using a smart pointer can we iterate through a container?
Ans: Yes. A container is a collection of elements or objects. It helps to properly
organize and store the data. Stacks, linked lists, arrays are examples of
containers. Following program shows how to iterate through a container using a
smart pointer.
#include
class smartpointer
{
private :
int *p ; // ordinary pointer
public :
smartpointer ( int n )
{
p = new int [ n ] ;
int *t = p ;
for ( int i = 0 ; i <= 9 ; i++ )
*t++ = i * i ;
}
int* operator ++ ( int )
{
return p++ ;
}
int operator * ( )
{
return *p ;
}
} ;
void main( )
{
smartpointer sp ( 10 ) ;
for ( int i = 0 ; i <= 9 ; i++ )
cout << *sp++ << endl ;
}
Here, sp is a smart pointer. When we say *sp, the operator * ( ) function gets
called. It returns the integer being pointed to by p. When we say sp++ the
operator ++ ( ) function gets called. It increments p to point to The next
element in the array and then returns the address of this new location.
43. Can objects read and write themselves?
Ans: Yes! This can be explained with the help of following example:
#include
#include
class employee
{
private :
char name [ 20 ] ;
int age ;
float salary ;
public :
void getdata( )
{
cout << "Enter name, age and salary of employee : " ;
cin >> name >> age >> salary ;
}
void store( )
{
ofstream file ;
file.open ( "EMPLOYEE.DAT", ios::app | ios::binary ) ;
file.write ( ( char * ) this, sizeof ( *this ) ) ;
file.close( ) ;
}
void retrieve ( int n )
{
ifstream file ;
file.open ( "EMPLOYEE.DAT", ios::binary ) ;
file.seekg ( n * sizeof ( employee ) ) ;
file.read ( ( char * ) this, sizeof ( *this ) ) ;
file.close( ) ;
}
void show( )
{
cout << "Name : " << name
<< endl << "Age : " << age
<< endl << "Salary :" << salary << endl ;
}
} ;
void main( )
{
employee e [ 5 ] ;
for ( int i = 0 ; i <= 4 ; i++ )
{
e [ i ].getdata( ) ;
e [ i ].store( ) ;
}
for ( i = 0 ; i <= 4 ; i++ )
{
e [ i ].retrieve ( i ) ;
e [ i ].show( ) ;
}
}
Here, employee is the class whose objects can write and read themselves. The
getdata( ) function has been used to get the data of employee and store it in the
data members name, age and salary. The store( ) function is used to write an
object to the file. In this function a file has been opened in append mode and
each time data of current object has been stored after the last record (if any) in
the file. Function retrieve( ) is used to get the data of a particular employee from
the file. This retrieved data has been stored in the data members name, age and
salary. Here this has been used to store data since it contains the address of the
current object. The function show( ) has been used to display the data of
employee.
44. Why is it necessary to use a reference in the argument to the copy
constructor?
Ans : If we pass the copy constructor the argument by value, its copy would get
constructed using the copy constructor. This means the copy constructor would
call itself to make this copy. This process would go on and on until the compiler
runs out of memory. This can be explained with the help of following example:
class sample
{
int i ;
public :
sample ( sample p )
{
i = p.i ;
}
} ;
void main( )
{
sample s ;
sample s1 ( s ) ;
}
While executing the statement sample s1 ( s ), the copy constructor would get
called. As the copy construct here accepts a value, the value of s would be
passed which would get collected in p. We can think of this statement as sample
p = s. Here p is getting created and initialized. Means again the copy constructor
would get called. This would result into recursive calls. Hence we must use a
reference as an argument in a copy constructor.
45.
46. Virtual Multiple Inheritance:
A class b is defined having member variable i. Suppose two classes d1 and d2
are derived from class b and a class multiple is derived from both d1 and d2. If
variable i is accessed from a member function of multiple then it gives error as
'member is ambiguous'. To avoid this error derive classes d1 and d2 with
modifier virtual as shown in the following program.
#include
class b
{
public :
int i ;
public :
fun( )
{
i = 0 ;
}
} ;
class d1 : virtual public b
{
public :
fun( )
{
i = 1 ;
}
} ;
class d2 : virtual public b
{
public :
fun( )
{
i = 2 ;
}
} ;
class multiple : public d1, public d2
{
public :
fun( )
{
i = 10 ;
}
} ;
void main( )
{
multiple d ;
d.fun( ) ;
cout << d.i ;
}
46. Can we use this pointer in a class specific, operator-overloading function for
new operator?
Ans: No! The this pointer is never passed to the overloaded operator new()
member function because this function gets called before the object is created.
Hence there is no question of the this pointer getting passed to operator new( ).
47. Can we allocate memory dynamically for a reference?
Ans: No! It is not possible to allocate memory dynamically for a reference. This
is because, when we create a reference, it gets tied with some variable of its
type. Now, if we try to allocate memory dynamically for a reference, it is not
possible to mention that to which variable the reference would get tied.
48. When should I overload new operator on a global basis or a class basis?
Ans: We overload operator new in our program, when we want to initialize a
data item or a class object at the same place where it has been allocated
memory. The following example shows how to overload new operator on global
basis.
#include
#include
void * operator new ( size_t s )
{
void *q = malloc ( s ) ;
return q ;
}
void main( )
{
int *p = new int ;
*p = 25 ;
cout << *p ;
}
When the operator new is overloaded on global basis it becomes impossible to
initialize the data members of a class as different classes may have different
types of data members. The following example shows how to overload new
operator on class-by-class basis.
#include
#include
class sample
{
int i ;
public :
void* operator new ( size_t s, int ii )
{
sample *q = ( sample * ) malloc ( s ) ;
q -> i = ii ;
return q ;
}
} ;
class sample1
{
float f ;
public :
void* operator new ( size_t s, float ff )
{
sample1 *q = ( sample1 * ) malloc ( s ) ;
q -> f = ff ;
return q ;
}
} ;
void main( )
{
sample *s = new ( 7 ) sample ;
sample1 *s1 = new ( 5.6f ) sample1 ;
}
Overloading the operator new on class-by-class basis makes it possible to
allocate memory for an object and initialize its data members at the same place.
49. How would you define a pointer to a data member of the type pointer to
pointer?
Ans: The following program demonstrates this...
#include
class sample
{
public :
sample ( int **pp )
{
p = pp ;
}
int **p ;
} ;
int **sample::*ptr = &sample::p ;
void main( )
{
int i = 9 ;
int *pi = &i ;
sample s ( &pi ) ;
cout << ** ( s.*ptr ) ;
}
Here, ptr is the pointer to data member p of class sample, which in turn is a
pointer pointing to an int.
50. How do I write a code to catch multiple types of exceptions in one single
catch block?
Ans: The following program demonstrates the use of a single catch block to
catch multiple exceptions.
#include
class test
{
} ;
class sample
{
public :
void fun1( )
{
throw 99 ;
}
void fun2( )
{
throw 3.14f ;
}
void fun3( )
{
throw "error" ;
}
void fun4( )
{
throw test( ) ;
}
} ;
void main( )
{
try
{
sample s ;
s.fun4( ) ;
s.fun1( ) ;
s.fun2( ) ;
s.fun3( ) ;
}
catch ( ... )
{
cout << "strange" ;
}
}
Here, different types of exceptions are thrown by the member functions of the
class sample. While catching the exception instead of four different catch blocks
we can as well define one single catch block. Note the syntax for defining the
catch block, where we have used three dots (?) in the formal parameter list. This
indicates that any thrown exception should get caught in the same catch block.
When the exception is thrown from the fun4( ) control reaches the catch block,
ignoring the rest of the calls.
51. Can we return an error value from the constructor of a class?
Ans: No. We cannot return any error value from the constructor, as the
constructor doesn't have any return type. However, by throwing an exception we
can pass value to catch block. This is shown in the following example:
#include
class sample
{
public :
sample ( int i )
{
if ( i == 0 )
throw "error" ;
}
} ;
void main( )
{
try
{
sample s ( 0 ) ;
}
catch ( char * str )
{
cout << str ;
}
}
In this program, the statement throw "error" ; would throw an exception when
an object s of the class sample would get created. The catch block would collect
the string error.
52. How do I define the member function of a template class, which has to be
defined outside the template class. The function receives an object of its own
class as a parameter and returns the value of the same type.
Ans: The following example shows how we can define such a function.
sample sample::fun ( sample s )
{
// code
}
Here, the first sample indicates the return type of the function and the next
sample is used for the scope of function.
53. How name mangling can be prevented?
Ans: To avoid name mangling the function should be declared with an extern "C"
attribute. Functions declared as extern "C" are treated as C-style functions.
Hence the compiler does not mangle them. The following code snippet shows
how to declare such a function.
#include
extern "C" void display( )
{
cout << "See the effect of C in C++ " ;
}
void main( )
{
display( ) ;
}
54. Can we allocate memory dynamically for a reference?
Ans: No, it is not possible to allocate memory dynamically for a reference. A
reference is initialized at the time of creation. Trying to allocate memory
dynamically for a reference creates a problem in initializing it. Thus, the compiler
does not allow us to dynamically allocate the memory for references.
55. What is RTTI?
Ans: RTTI stands for 'Run Time Type Information'. We use virtual function
mechanism where we can call derived class's member functions using base
class's pointer. However, many times we wish to know the exact type of the
object. We can know the type of the object using RTTI. A function that returns
the type of the object is known as RTTI functions. C++ supports two ways to
obtain information about the object's class at run time, they are typeid( )
operator and dynamic_cast operator.
56. What is Data Conversion?
Ans: Assignments between types whether they are basic or user-defined, are
handled by the compiler. If the variables are of different basic types compiler
calls a special routine to convert the value. But if we want to convert between
user-defined data type and basic types we have to write conversion routine
ourselves. A conversion routine to convert user-defined data type string to
integer is shown below:
class string
{
private :
char str[20] ;
public :
string( )
{
}
string ( char *s )
{
strcpy ( str, s ) ;
}
operator int( )
{
return 123 ; // Write logic to convert string to integer
}
} ;
main( )
{
string s2 = "123" ;
int i1 = int ( s2 ) ;
cout << endl << i1 ;
}
57. How to obtain type information using typeid( ) operator?
Ans: typeid( ) operator takes an object, a reference or a pointer and returns its
type. Following program shows how to use the typeid( ) operator.
#include
#include
class Base
{
public :
virtual void show( )
{
}
};
class Der1 : public Base
{
} ;
void main( )
{
Base *b1 ;
cout << endl << typeid ( b1 ).name( ) ;
Der1 d1 ;
b1 = &d1 ;
cout << endl << typeid ( *b1 ).name( ) ;
cout << endl << typeid ( 12 ).name( ) << endl << typeid ( 12.5 ).name( ) ;
}
The output of this program will be
Base*
Der1
int
double
RTTI operators must be used for polymorphic class (class having virtual
function) only. For non-polymorphic class static type information is returned.
58. How to use RTTI with class templates?
Ans: Templates can generate different classes. We may wish to get the type of
class, which we are working in. The following program shows how to use RTTI
operator typeid( ) with class template.
#include
#include
template
class base
{
public :
base( )
{
cout << typeid ( *this ).name( ) << "Constructor" << endl ;
}
T add ( T a, T b )
{
return a + b ;
}
~base( )
{
cout << typeid ( *this ).name( ) << "Destructor" << endl ;
}
} ;
void main( )
{
base b1 ;
cout << b1.add ( 10, 20 ) << endl ;
base b2 ;
cout << b2.add ( 5.5, 10.5 ) << endl ;
}
59. We can use following C++ operators for typecasting. static_cast is used for
castless conversions, narrowing conversions, conversion from void* and implicit
type conversions. const_cast is used to convert a const to a non-const.
reinterpret_cast is used to assign one kind of pointer to another.
60. What will be the output of the following program?
#include
class A
{
public :
A( )
{
cout << "Reached in Constructor\n" ;
}
} ;
void main( )
{
A a( ) ;
A b ;
}
Output : Reached in Constructor
Constructor gets called only once when the object b is created. When the
statement A a( ) ; gets executed constructor does not get called. This is because
compiler takes this statement as a prototype declaration of function a( ) that
returns an object of class A. However, if we pass arguments like
A a ( 10 ) ;
Compiler would search for one argument constructor and if not found would flash
an error.
61. What is a container?
Ans: A container is an object that holds other objects. Various collection classes
like List, Hash Table, AbstractArray, etc. are the examples of containers. We can
use the classes to hold objects of any derived classes. The containers provide
various methods using which we can get the number of objects stored in the
container and iterate through the objects stored in it.
62. Function template overloading
One can declare several function templates with the same name and even
declare a combination of function templates and ordinary functions with the
same name. When an overloaded function is called, overload resolution is
necessary to find the right function or template function to invoke.
For example:
template < class T > T sqrt ( T ) ;
template < class T > complex < T > sqrt ( complex < T > ) ;double sqrt (
double ) ;
void f ( complex < double > z )
{
sqrt ( 2 ) ; // sqrt < int > ( int )
sqrt ( 2.0 ) ; // sqrt ( double )
sqrt ( z ) ; // sqrt < complex < double > ( complex < double > )
}
In the same way that a template function is a generalization of the notion of a
function, the rules for resolution in the presence of function templates are
generalizations of the function overload resolution rules. Basically, for each
template we find the specialization that is best for the set of function arguments.
Then we apply the usual function overload resolution rules to these
specializations and all ordinary functions.
63. Exception Handling in C++
In C++ we can handle run-time errors generated by c++ classes by using three
new keywords: throw, catch, and try. We also have to create an exception class.
If during the course of execution of a member function of this class a run-time
error occurs, then this member function informs the application that an error has
occurred. This process of informing is called 'throwing' an exception. The
following code shows how to deal with exception handling.
class sample
{
public :
class errorclass
{
} ;
void fun( )
{
if ( some error occurs )
throw errorclass( ) // throws exception
}
} ;
//application
void main( )
{
try
{
sample s ;
s.fun( ) ;
}
catch ( sample::errorclass )
{
// do something about the error
}
}
64. Consider the following code:
#include
class base
{
public :
int data ;
} ;
class d1 : public base
{
} ;
class d2 : public base
{
} ;
class der : public d1, public d2
{
public :
void showdata( )
{
cout << data ;
}
} ;
void main( )
{
der d ;
d.showdata( ) ;
}
If you run this program it is bound to give you errors. This is because of the
rules of inheritance:
1. Each base class not specified virtual will have its own sub-object representing
it. In the above program, if we create object of d1 it will have a sub-object of
class base containing a data member data. If we create an object of class der it
will have sub-objects of classes d1 and d2 and both the sub-objects will refer to
a separate copy of data. Hence, to access data from class der we will have to
mention the class name. For example, d1::data or d2::data.
2. If we want that only one sub-object should exist we must use the concept of
virtual base class. The single object of this will represent every base class of
given name that is specified to be virtual class. After making d1 and d2 as
virtual base class if we create an object of der only one sub-object would exist
and so accessing data would no longer give us errors.
65. How to declare a pointer to a member function?
Ans: Suppose, I wish to declare a pointer to a member function that receives an
int and returns an int. I will have to declare it as int (A::* ) ( int ). Following is
an example.
#include
class A
{
public :
int fun ( int f )
{
cout << "in fun\n" ;
return f * f ;
}
} ;
typedef int ( A:: *pfun ) ( int ) ;
void main( )
{
pfun p = A::fun ;
A a ;
int s = ( a.*p ) ( 6 ) ;
cout << s ;
}
66. What is the disadvantage of a template function?
Ans: A template function cannot be distributed in the obj form. This is because,
with which parameters the template function is going to be called is decided at
the run time only. Therefore an obj form of a template function cannot be made
by merely compiling it.
67. How to declare a pointer to the data members of a class?
Ans: Following program shows how to declare a pointer to non-function
members of a class.
#include
class A
{
public :
int a ;
void print( )
{
cout << a ;
}
} ;
void main( )
{
int A::*pa = &A::a ;
A obj ;
obj.*pa = 20 ;
obj.print( ) ;
}
Here, we have initialised the data member a using the pointer pa.
68. How to allocate memory for a multidimensional array dynamically?
Ans: Many times we need to allocate memory for a multidimensional array
dynamically. Because of complexity of pointers many find this difficult. Following
program allocates memory for a 3 x 3 array dynamically, copies contents of a 3
x 3 array in it and prints the contents using the pointer.
#include
#include
int a[ ][3] = {
1, 2, 3,
4, 5, 6,
7, 8, 9
} ;
void main( )
{
int **p ;
p = new int *[3] ;
for ( int i = 0 ; i < 3 ; i++ )
p[i] = new int[3] ;
for ( i = 0 ; i < 3 ; i++ )
for ( int j = 0 ; j < 3 ; j++ )
p[i][j] = a[i][j] ;
for ( i = 0 ; i < 3 ; i++ )
{
for ( j = 0 ; j < 3 ; j++ )
cout << p[i][j] ;
cout << "\n" ;
}
}
69. When should we use the :: ( scope resolution ) operator to invoke the virtual
functions?
Ans: Generally, :: operator is used to call a virtual function from constructor or
destructor. This is because, if we call a virtual function from base class
constructor or destructor the virtual function of the base class would get called
even if the object being constructed or destroyed would be the object of the
derived class. Thus, whenever we want to bypass the dynamic binding
mechanism we must use the :: operator to call a virtual function.
70. How do I use operators .* and ->* in a program?
Ans: The following code snippet demonstrates the use of .* and ->* operators.
#include
class sample
{
public :
int i ;
void fun( )
{
cout << "fun" << endl ;
}
} ;
void ( sample::*pf )( ) = &sample::fun ;
int sample::*pdm = &sample::i ;
void main( )
{
sample s ;
sample *p = new sample ;
( s .* pf )( ) ;
( p ->* pf )( ) ;
s .* pdm = 1 ;
p ->* pdm = 2 ;
cout << s .* pdm << endl ;
cout << p ->* pdm << endl ;
}
In the above program pf is a pointer to a function fun( ) of class sample, and
pdm is a pointer to a data member i of the same class sample. The object s of
the class sample is created statically. Next, p is a pointer to an object created
dynamically. The using the operator .* and ->* the member functions are called
and also the public data member is accessed.
71. What happens when we add an int value to a user defined type of object?
Ans: Whenever an int value is added to an object of user defined type, the
object would search for an overloaded operator int( ). This operator must be
defined in such a way that it always returns an int value. However, we need not
specify the return type as on doing so the compiler flashes an error.
#include
class sample
{
int i ;
public :
sample ( )
{
i = 10 ;
}
operator int( )
{
return this -> i ;
}
} ;
void main( )
{
sample s ;
int i ;
i = s + 10 ;
cout << i ;
}
In the above program on adding 10 to an object s, the value of i would become
20.
72. Can we have a reference to an array?
Ans: Yes, we can have a reference to an array.
int a[ ] = { 8, 2, 12, 9 } ;
int ( &r ) [ 4 ] = a ; // reference to an array
Here, r is a reference to an array of four elements. We can even print the
elements of array with the help of reference. This is shown in the following code
segment:
for ( int i = 0 ; i < 4 ; i++ )
cout << r [i] << endl ;
73. When friend function becomes indispensable...
Ans: Consider the following program.
#include
class distance
{
private :
int feet ;
public :
distance( )
{
feet = 0 ;
}
distance ( int f )
{
feet = f ;
}
distance operator + ( distance x )
{
int f = feet + x.feet ;
return distance ( f ) ;
}
} ;
void main( )
{
distance d1 ( 20 ), d2, d3 ;
d2 = d1 + 10 ;
d3 = 10 + d2 ;
}
If you run this program it is bound to give errors. The error lies in the statement
d3 = 10 + d2 ; We may think that since we have overloaded + operator this
statement would add 10 to d2. But this does not happen. This is because the
specified statement will get converted as d3 = 10.operator+ ( d2 ) ; This means
that this statement should call the operator+( ) function that takes an object of
distance class as parameter written in the float class, which is not possible. The
solution is to write operator+( ) as a 'friend' function. Declare operator+ function
in distance class as given below:
friend distance operator + ( distance x1, distance x2 ) ;
and define it outside the class as shown below:
distance operator + ( distance x1, distance x2 )
{
int f = x1.feet + x2.feet ;
return distance ( f ) ;
}
When compiler would see that the 'friend' operator+( ) function is available it
would convert the statement d3 = 10 + d2 as operator+ (10, d2 ). Now since 10
is passed as a parameter not as a calling object there would be no error. Thus in
such cases 'friend' function becomes indispensable.
74. How to use a memory as a stream?
Ans: Suppose, details of an employee such as name, designation, age, etc. are
stored in different types of variables. Now, if we wish to concatenate these
details in a character array we will have to use various string manipulation
functions like strcpy( ) and strcat( ). Instead of using these functions we can use
more easy and clean way to gather the details in the char array in the form of
streams. We can declare the memory allocated for the array as stream and use
the << operator to store variables having different types in this memory.
Following program shows how to achieve this.
#include
void main( )
{
char buff [50] ;
char str[ ] = "Sanjay" ;
char desig[ ] = "Manager" ;
char jd[ ] = "27/12/1995" ;
int age = 35 ;
ostrstream o ( buff, sizeof ( buff ) ) ;
o << str << endl << desig << endl << jd << endl << age << ends ;
cout << buff ;
}
As shown in the program we can also use the manipulators and formatting flags.
The output of this program will be:
Sanjay
Manager
27/12/1995
35
75. How would you declare and initialize reference to a data member?
Ans: Sometimes we may need to declare a data member, which is a reference to
another data member of the class as shown below:
class A
{
public :
char *p ;
char *&rp ;
} ;
We can't initialize a reference to a data member at the time of declaration. It
should be initialized using 'member wise initialization as shown below.
#include
class A
{
public :
char *p ;
char *&rp ;
A( ) : rp ( p )
{
p = "" ;
}
A ( char *s ) : rp ( p )
{
p = s ;
}
} ;
void main( )
{
A a ( "abcd" ) ;
cout << a.rp ;
}
76. iostream library has made it easy to read data from various input devices
and write data to the output devices. The following program shows how to print
a disk file 'data.dat' on the printer using stream classes. Every hardware device
has a familiar name given by the operating system. The printer is generally
connected to the first parallel port. So, the file name for the printer should be
PRN or lpt1.
#include
void main( )
{
ifstream i ( "data.dat" ) ;
ofstream o ;
o.open ( "PRN" ) ;
char ch ;
while ( 1 )
{
i.get ( ch ) ;
if ( i.eof( ) )
break ;
o.put ( ch ) ;
}
o.put ( '\x0C' ) ;
}
77. We know that a destructor is automatically called when an object of a class
goes out of scope. There is another case where destructor is called
automatically. If an object is created in a try block and an exception is thrown
after the object is created, then the destructor is called automatically.
78. Can a function call be at the left hand side of the assignment operator?
Ans: Yes. Following program shows how it is possible.
#include
class ref
{
private :
struct data
{
int a ; char *p ;
} d1, d2 ;
public :
data &set ( )
{
return d1 ;
}
data &get ( )
{
cin >> d2.a >> d2.p ;
return d2 ;
}
} ;
void main( )
{
ref r ;
r.set( ) = r.get( ) ;
r.print( ) ;
}
In the above program the functions get( ) and set( ) both return a reference to
the object of the structure data. We have assigned the reference returned by
get( ) to the reference returned by set( ) function. That is, we are assigning d2
to d1. So, the values of d2 would get assigned to d1. You can check this out by
printing the values of d1.
79. If a class contains a virtual function a pointer called VPTR is created. This
VPTR becomes a part of every object of that class. The first two bytes (in DOS)
are occupied by VPTR. We can prove this by displaying the first two bytes of
memory allocated for the objects. Following program shows how this can be
achieved.
#include
class vir
{
public :
virtual void f( )
{
}
} ;
void main( )
{
vir v, v1 ;
int *p1 = ( int* ) &v ;
int *p2 = ( int* ) &v1 ;
cout << endl << *p1 << " " << *p2 ;
}
80. Exception Handling in C++
In C++ we can handle run-time errors generated by c++ classes by using three
new keywords: throw, catch, and try. We also have to create an exception class.
If during the course of execution of a member function of this class a run-time
error occurs, then this member function informs the application that an error has
occurred. This process of informing is called 'throwing' an exception. The
following code shows how to deal with exception handling.
class sample
{
public :
class errorclass
{
} ;
void fun( )
{
if ( some error occurs )
throw errorclass( ) // throws exception
}
} ;
//application
void main( )
{
try
{
sample s ;
s.fun( ) ;
}
catch ( sample::errorclass )
{
// do something about the error
}
}
81. Accessing a private data member from a different Object...Different objects
of the same class can access each other's members, even if these members are
private. For example:
#include < iostream.h >
class sample
{
float f ;
public :
sample ( float ff )
{
f = ff ;
}
void fun ( sample* objptr )
{
objptr -> n = 0 ;
cout << "Value of this objects f is : " << f << endl ;
cout << "Value of other objects f" << objptr -> n << endl ;
} // another object's private member!
} ;
void main( )
{
sample s1 ( 6.5f ) , s2 ( 2.5f ) ;
s1.f ( &s2 ) ; // s1 changes s2's n
}
Typically, this coding style should be avoided. However, you should be aware
that private members of an object can be changed by another object of the
same
type. Therefore, in certain special conditions, this coding style may be useful.
82. Can you access private data members of a class from out side the class?
Ans: Yes. This program shows how.
#include
class emp
private :
int i ;
public :
emp( )
{
i = 10 ;
}
} ;
void main( )
emp *p = new emp ;
int *pi = (int*) p ;
cout << *pi ;
*pi = 20 ;
cout << *pi ;
}
The pointer to the class is typecasted in an integer pointer. With the help of this
pointer private data member 'i' is accessed in main( ).
83. Why creating array of references is not possible?
Ans: The array name always refers or points to the zeroeth element. If array is
of references then the array name would point to the zeroeth element which
happens to be a reference. Creating pointer to a reference is not valid. So,
creating array of references too is not possible.
84. How do I call a virtual function of a class using a pointer to a function ?
Ans :
#include
class Cvirtual
{
public :
virtual float vfun( )
{
cout << "from vfun" << endl ;
return 7.03f ;
}
} ;
void main( )
{
Cvirtual obj ;
int **p = ( int ** ) &obj ;
float ( *pf1 ) ( ) ;
pf1 = ( float ( * ) ( ) ) **p ;
float f = ( *pf1 ) ( ) ;
cout << "return val = " << f << endl ;
}
In the above program class Cvirtual consists of a virtual function vfun(). In
variable p we have stored the address of an object of class Cvirtual. While doing
so, we have type casted the address of obj to int **, because obj holds a hidden
data member called vptr, which in turn holds the address of virtual function
vfun( ). In pf1, a pointer to a function, we are collecting the address of the
virtual function vfun( ). Thus the value returned by vfun( ) would then get
collected in f.
85. Why an overloaded new operator defined in a class is static?
Ans: An overloaded new function is by default static even if it is not declared so.
This is because non-static member functions can be called through an object
only. But when an overloaded new operator function gets called the object
doesn't stand created. Since new operator function itself is responsible for
creating the object. Hence to be able to call a function without an object, the
function must be static.
86. What is a pure virtual destructor?
Ans: Like a pure virtual function we can also have a pure virtual destructor. If a
base class contains a pure virtual destructor it becomes necessary for the
derived classes to implement the destructor. An ordinary pure virtual function
does not have a body but pure virtual destructor must have a body. This is
because all the destructors in the hierarchy of inheritance are always called as a
part of destruction.
87. When we are required to find offset of an element within a structure? or,
how do we call the function of an outer class from a function in the inner class?
(The inner class is nested in the outer class)
Ans:
#include
class outer
{
int i ;
float f ;
public :
class inner
{
public :
infunc( )
{
outer *pout ;
pout = (outer*) this - ( size_t ) &( ( ( outer* ) 0 ) -> in ) ;
pout -> outfunc( ) ;
}
} ;
inner in ;
outfunc( )
{
cout << "in outer class's function" ;
}
} ;
void main( )
{
outer out ;
out.in.infunc( )
}
In the above example we are calling outer::outfunc( ) from inner::infunc(). To
call outfunc( ) we need a pointer to the outer class. To get the pointer we have
subtracted offset of the inner class's object (base address of outer class's object
- address of inner class's object) from address of inner class's object.
88. void f ( float n, int i = 10 ) ;
void f ( float a ) ;
void main( )
{
f ( 12.6 ) ;
}
void f ( float n, int i )
{
}
void f ( float n )
{
}
The above program results in an error (ambiguous call) since without the default
argument the two functions have arguments that are matching in number, order
and type.
89. Some programs need to exercise precise control over the memory areas
where data is placed. For example, suppose we wish to read the contents of the
boot sector into a structure. For this the byte arrangement of the structure
elements must match the arrangement of various fields in the boot sector of the
disk.
The #pragma pack directives offer a way to fulfill this requirement. The
#pragma pack directive specifies packing alignment for structure and union
members. The #pragma takes effect at the first structure or union declaration
after the #pragma is seen. Consider the following structure:
#pragma pack (1)
struct emp
{
int a ;
float s ;
char ch ;
} ;
#pragma pack( )
Here, #pragma pack ( 1 ) lets each structure element to begin on a 1-byte
boundary. Hence the size of the structure will be 9. (int - 4, float - 4, char - 1).
If we use #pragma pack ( 2 ) each structure element can begin on a 2-byte
boundary. Hence the size of the structure will be 10. (int - 4, float - 4, char - 2).
90. How to restrict a friend class's access to the private data members?
Ans: If we declare a class as a friend of our class the friend class can access the
private data members of our class. However, if we want we can restrict this
access to some selective functions of the class. Following program shows how to
achieve this:
#include
class X
{
public :
void print ( class Z &z ) ;
} ;
class Z
{
private :
int i ;
public :
Z ( int ii )
{
i = ii ;
}
friend X::print ( class Z &z ) ;
} ;
void X::print ( Z &z1 )
{
cout << z1.i ;
}
main( )
{
Z z ( 10 ) ;
X x ;
x.print ( 10 ) ;
}
In the above program only the X::print( ) function can access the private data
members of class Z.
91. What is name mangling?
Ans: C++ enables you to assign the same function name to more than one
functions but with different parameter types. This feature is called function
overloading. But when we give several functions the same name, how does the
compiler decide which particular function is to be called? C++ solves this
problem by applying a process called name mangling. Name mangling applies a
decorated name to the function. The mangled name includes tokens that identify
the functions' return type and the types of its arguments.
class test
{
public :
void fun ( int a, char b ) ;
void fun ( char *c, float y ) ;
} ;
void main( )
{
test s1 ;
s1.fun ( 65, 'A' ) ;
s1.fun ( "Anil", 5.5f ) ;
}
At the time of resolving the calls to fun( ) function the linker would not be able
to find the definition of the overloaded function fun( ) and it would report an
error. If you look at these errors you will see the mangled names like,
(?fun@test@@QAEXJJ@Z) and (?fun@test@@QAEXMM@Z). Note that different
compilers may use different name mangling schemes.
92. How would you call a C function from C++ code?
Ans: Using extern "C".
The function prototype must be preceded by extern "C". More than one C
functions can be grouped inside braces as shown below:
extern "C"
{
void f( ) ;
void f1( ) ;
}
// In cfunc.c
#include
void f( )
{
printf ( "in f( )" ) ;
}
// In func.cpp
#include
extern "C" void f( ) ;
void main( )
{
f( ) ;
}
Ensure that both .c and .cpp files are in the same project.
93. How to restrict the number of floating-point digits displayed ?
Ans: When we display floating-point values, we can use the setprecision
manipulator to specify the desired number of digits to the right of the decimal
point.
For example,
cout << setprecision ( 3 ) << 12.34678 ;
This statement would give the output as 12.347.
94. What is a wild pointer ?
Ans: A wild pointer is the one that points to a garbage value. For example, an
uninitialized pointer that contains garbage value or a pointer that refers to
something that no longer exists.
95. How friend function helps to increase the versatility of overloaded operators?
Ans: Consider the following statement,
s2 = s1 * 2 ;
where, s1 and s2 are objects of sample class. This statement would work if the
overloaded operator * ( sample s ) or conversion function is provided in the
class. Internally this statement would get converted to,
s2 = s1.operator * ( 2 ) ;
The function materializes because it is called with an object s1. The this pointer
of s1 would get passed implicitly. To collect 2 in s, first the compiler would call
the one-argument constructor, then it would build a nameless object, which then
would get collected in s. However, if we write the above statement as,
s2 = 2 * s1 ;
then it won't compile. This is because the call now would get treated as,
s2 = 2.operator * ( s1 ) ;
and 2 is not an object. The friend function helps to get rid of such a situation.
This is shown in the following program.
#include
class sample
{
private :
int i ;
public :
sample ( int ii = 0 )
{
i = ii ;
}
void showdata( )
{
cout << i << endl ;
}
friend sample operator * ( sample, sample ) ;
} ;
sample operator * ( sample s1, sample s2 )
{
sample temp ;
temp.i = s1.i * s2.i ;
return ( temp ) ;
}
void main( )
{
sample s1 ( 10 ), s2 ;
s2 = s1 * 2 ;
s2.showdata( ) ;
s1 = 2 * s2 ;
s1.showdata( ) ;
}
Here the operator *( ) function takes two parameters. This is because the
operator function is no longer a member function of the class. It is a friend of
the class sample. Thus the statement s2 = s1 * 2 ; would not take the form
s2.operator * ( 2 ). This example shows that using friend permits the overloaded
operators to be more versatile.
96. What is a const_cast?
Ans: The const_cast is used to convert a const to a non-const. This is shown in
the following program.
#include
void main( )
{
const int a = 0 ;
int *ptr = ( int * ) &a ; // one way
ptr = const_cast ( &a ) ; // better way
}
Here, the address of the const variable a is assigned to the pointer to a non-
const variable. The const_cast is also used when we want to change the data
members of a class inside the const member functions. The following code
snippet shows how to do this.
class sample
{
private :
int data ;
public :
void fun( ) const
{
( const_cast ( this ) ) -> data = 70 ;
}
} ;
97. Using a smart pointer we can make an object appear like a pointer.
If a class overloads the operator -> then any object of that class can appear like
a pointer when the operator -> ( ) is called. The following program illustrates
this.
#include
class test
{
public :
void fun( )
{
cout << "fun of smart pointer" ;
}
} ;
class smartpointer
{
test t ;
public :
test* operator ->( )
{
return &t ;
}
} ;
void main( )
{
smartpointer sp ;
sp -> fun( ) ;
}
The beauty of overloading operator -> is that even though sp is an object we
can make it work like a pointer. The operator -> ( ) returns the address of the
object of the type test. Using this address of the test object the function fun( ) of
the class test gets called. Thus even though fun( ) is not a member of
smartpointer class we can still call it using sp.
98. Can we apply delete on this pointer inside a member function?
Ans : Yes! If the member function of a class is called using a pointer to an
object, which is allocated dynamically, the object would get deleted. But if the
member function is called using the object, which is allocated statically, then a
runtime error would occur. This is because we cannot call delete on statically
allocated objects. This is illustrated in the following example.
class sample
{
private :
int i ;
public :
void fun( )
{
delete this ;
}
} ;
void main( )
{
sample *s = new sample ;
s -> fun( ) ; // no error
sample s1 ;
s1.fun( ) ; // would throw a runtime error
}
99. Why can't data members of a class be initialized at the time of declaration as
given in the following code?
class emp
{
private :
int j = 10 ;
} ;
Ans: Memory for data members of a class is allocated only when object of
that class is created. One cannot store data in a memory location, which does
not exist at all. Therefore initialization at the time of declaration is not possible.
100. Why in a copy constructor an object is collected in a reference to object as
shown below?
#include
class emp
{
public :
emp( )
{
}
emp ( emp& )
{
cout << "copy" ;
}
} ;
void main( )
{
emp e ;
emp e1 = e ;
}
Ans: A copy constructor is called when an object is created and initialised at the
same time. It is also called when object is passed to a function. So, If we pass
the object to copy constructor copy constructor would get called recursively.
Thus it will stuck up in an infinite loop.
101. What is Early Binding and Dynamic Binding?
Ans: The term binding refers to the connection between a function call and the
actual code executed as a result of the call. Early Binding: If which function is to
be called is known at the compile-time it is known as static or early binding.
Dynamic Binding: If which function is to be called is decided at run time it is
called as late or dynamic binding. Dynamic binding is so called because the
actual function called at run-time depends on the contents of the pointer. For
example, call to virtual functions, call to functions to be linked from dlls use late
binding.
102. When can we use the function ostrstream::freeze( )?
Ans: While outputting data to memory in the in-memory formatting we need to
create an object of the class ostrstream. The constructor of ostrstream receives
the address of the buffer but if we want that the ostrstream object should do its
own memory management then we need to create an ostrstream object with no
constructor arguments as:
ostrstream s ;
Now s will do its own memory management. We can stuff as many bytes into it
as we want. If it falls short of memory, it will allocate more memory. If it cannot,
it may even move the block of memory. When the object goes out of scope, the
heap storage is automatically released. This is a more flexible approach if we do
not know how much space we are going to need. If we want the physical
address of the memory used by s we can obtain it by calling the str( ) member
function:
char* p = s.str( ) ;
Once str( ) has been called then the block of memory allocated by ostrstream
cannot be moved. This is logical. It can't move the block since we are now
expecting it to be at a particular location. In such a case we say that ostrstream
has freezed itself. Once frozen we can't add any more characters to it. Adding
characters to a frozen ostrstream results in undefined behavior. In addition, the
ostrstream is no longer responsible for cleaning up the storage.
. square root(x)/3 = 243/x value of x ???? ans. 81
2. qn from loss and gain what is the net profit ? ans. 25 percent
3.if + sign is represented as /, - as *(into,not power do not confuse) , * as - and / ! as + then what will be the value of (480+20*20)-16 /12 ??? ans. 76
4. which of the following will come next in the series IAR, GET , EIV , COX ans. AUZ
5. B is in DOUBT in the same way H is in choices.... HOPE , HOUSE , HONEST , INHIBIT ans. HOUSE(confirm it i am not sure)
6. if P+Q represents P is the sister of Q, P-Q represents P is the father of Q and P*Q represents P is the brother of Q then which will represent K is the aunt of N
ans. K+Q-M*N
7.In a class puspa ranks 8th from top and 28th from bottom.how many students are there in the class????? ans. 35
8. 23 percent of a no is 230, 4/23rd of the number is ?????? ans.none of these
9. sum of a number and its one third = 60, 4/5th
of the number is ????? ans. 36
10.if A to Z is written in the reverse order which is the 5th letter to the right of the 11th letter from the left???? ans . K
11.In a rectangle length = twice that of breadth , cost of! grounding is 60 rupees per meter,total cost is 3600/- what is the length in meter???????? ans. 20
12.qn on sequences
which will come next??
11 13 17 23 31 ? 53 67 ans. 41 13. 18 27 64 125 ? 343 512 ans. 216 14. 2 3 10 15 26 ? 50 63 ans.please find out 15. 0 2 10 30 68 ? - - (cannot recall) please find out
16.if simon is taller than A, A is shorter then gayatri, sharat is taller then............ ans. gayatri
17. If a 100m train crosses a platform of 75m in 7.5 secs,another train of 120m coming opposite, how much time it takes the first train to cross the second one .. ans. none of these
18. Telegram system, 1st 15 words in x rupees,next per letter in y rupees, if z > 15 what is the total cost of c??????? ans. x+(z-15)*y
19. -------------------------------- another st line cuts this in two angles, one is 4x+36 degs and another is y degs , what is the value of x????? ans. (144-y)/4
20. If 6 i! s preceded by 8 but not followed by 9 immediately,how many..................... ans. 2
21.in the letter PREDILECTION ,1st and 3rd is
swapped,so is 2nd and 4th,5th and 7th,6th and 8th what will be 8th letter from left??????? ans. L
22.If rup ho la represents anand is fat cup wir ho represents mina is tall la dup mil represents boys are fat and wir lar nap represents tall and fat
what reprens mina ?????? ans. cup
23.If SINDHU is written as VIEOJT,SURESH is written as?????? ans.none of these
24.if hit-----> one rupee gain and miss--------> one rupee loss,one can try 100 times,he got total 30/-, how many hit?????????? ans. 65
25.In a circle, G is to the left of M , V is to the right of M,R is to the right of V ,A is to the right of R ,G is to the right of A and M is to the right of G, swap the position of G and R,V and A then who will be to the left of M
ans. R
26.there was a histogram qn. given the! capitals of 5 companies for the academic year 1995-2000,find out
a> which company will increase its capital consistently??? b> Average capital in the yr 1998????? c> average of company 2nd????? d> which company increases its capital by 10% at 1998 from that of 1995 data were 42.0 in 1995 and 46.20 in 1998 this company is the correct answer..... e> what is the max. o/p in a yr,ans.. try to sum all the cols and find the max. this were easy,i can not remeber u have to do it fast.
27. there was a qn on flowchart,which was easy, i can not recall exactly.
28. in english u have to find out errors in 5 fragments of a sentence ( GRE pattrn) which were easy, 4 qns , and fill in the blanks which were extremely difficult(4 qns) read Barron's guide
also there was a long passage which attempt at last,this was in qn no from 16-20,if it is answer of 18 is choice 2, 19 is 2 and that of 20 is 3
******** i am sending t! he answer of some of the qns,the qn will be in order,do not fear test- 1 1- 5, 2- 1, 3-5, 4- 4 , 5 - 3 , 6 - 1 , 8 - 5 , 9 - 5 , 10 - 4 , 11 - 2 , 12 - 1 , 15 - 5 , 16 - 1 , 18 - 3 , 19 - 3 , 20 - 5 26- 1 , 17 - 4 , 28 - 5 , 29 - 3 , 30 - 3 ( qn no - choice)
test -3
1 - 3 , 2 - 4 , 3 - 2 , 4 - 2 , 5 - 2 , 6 - 1 , 7 - 2 , 8 -1 , 18 - 2 , 19 - 2 , 20 - 3
that's all folks!!!!!!! that's all i
memorise..... u have to do faster...... qns were
easy..........
all the very best for ur campus........... mail
bout ur result .........be in touch.........
Hi friends!
This is new HCL aptitude paper conducted in
Jadavpur University 24 th July 2001. After
qualifying this test u have to face technical
test.
OK GOOD LUCK.
Duration:30min check the answers
QUANTITATIVE APTITUDE
Q.NO.1&2
A FACTORY produces a product (measured in cubic
feet) over seven days as per the following
schedule:
Day1 Day2 Day3 Day4 Day5 Day6 Day! 7
150 180 120 250 160 120 150
The finished goods are to be transported to the
market by a truck having a capacity of 2000 cubic
feet. Any finished goods (ready at the end of the
day) retained overnight at the factory will incur
a storage cost of rs.5 per cubic foot for each
night of storage. The hiring cost for the truck
is rs.1000 per day.
1.If the total cost of transportation and storage
is to be minimized, the truck should operate on
(a) 2nd,4th,6thand 7th days
(b) only the 7th day
(c) 2nd,4th,5th and 7th days
(d) only on 4th and 5th days
(e) none of these
ans:( )
2.if the storage cost reduces to re.0.80per cubic
foot per night,the truck should operate on
(a)2nd,4th,5thand 7th days
(b)only the 7th day
(c)2nd,4th,5th and 7th days
(d)only on 4th and 5th days
(e)none of these
ans:( )
3.how many times do the hands of a clock contain
a right angle in a day?
(a)48 (b)44 (c)4 (d)8 (e)none of these ans:( )
! 4.1-1/3+1/3^2-1/3^3+^^^.to infinite
(a) 2/3 (b) 3/5 (c)3/4 (d) 4/5(e)none of these
ans:( c)
5. 5 students take a calculus test on which the
lowest possible score is 0.&the highest possible
score is 100.the
students,rahul,aparna,suzy,rohit& shitij, each
earned different scores.suzy outscored aparna by
8 points;rohit scored 1 times what rahul
scored&10 points higher than shitiji,who scored 2
points higher than aparna.the second highest
score was 89.what score did each student achieve
on the test? Rahul-------------
1. For a CMOS inverter, the transition slope of Vout vs Vin DC characteristics can be increased
(steeper transition) by:
a. Increasing W/L of PMOS transistor
b. Increasing W/L of NMOS transistor
c. Increasing W/L of both transistors by the same factor
d. Decreasing W/L of both transistor by the same factor
2. Minimum number of 2-input NAND gates that will be required to implement the function:
Y = AB
+ CD + EF is
a. 4
b. 5
c. 6
d. 7
3. Consider a two-level memory hierarchy system M1 & M2. M1 is accessed first and on miss
M2 is
accessed. The access of M1 is 2 nanoseconds and the miss penalty (the time to get the data
from
M2 in case of a miss) is 100 nanoseconds. The probability that a valid data is found in M1 is
0.97.
The average memory access time is:
a. 4.94 nanoseconds
b. 3.06 nanoseconds
c. 5.00 nanoseconds
d. 5.06 nanoseconds
4. Interrupt latency is the time elapsed between:
a. Occurrence of an interrupt and its detection by the CPU
b. Assertion of an interrupt and the start of the associated ISR
c. Assertion of an interrupt and the completion of the associated ISR
d. Start and completion of associated ISR
5. Which of the following is true for the function (A.B + A.C + B.C)
a. This function can glitch and can be further reduced
b. This function can neither glitch nor can be further reduced
c. This function can glitch and cannot be further reduced
d. This function cannot glitch but can be further reduced
6. For the two flip-flop configuration below, what is the relationship of the output at B to the
clock
frequency?
a. Output frequency is 1/4th the clock frequency, with 50% duty cycle
b. Output frequency is 1/3rd the clock frequency, with 50% duty cycle
c. Output frequency is 1/4th the clock frequency, with 25% duty cycle
d. Output frequency is equal to the clock frequency
XOR
A B
Q
D Q
D
CLK
CLK
Q
Q
7. The voltage on Node B is:
a. 0
b. 10
c. 10
d. 5
10O
10O 10O
+ +
10O
10O
10V 20V
_ _
GND B
8. A CPU supports 250 instructions. Each instruction op-code has these fields:
The instruction type (one among 250)
A conditional register specification
3 register operands
Addressing mode specification for both source operands
The CPU has 16 registers and supports 5 addressing modes. What is the instruction op-code
length in bits?
a. 32
b. 24
c. 30
d. 36
9. In the iterative network shown, the output Yn of any stage N is 1 if the total number of 1s
at the
inputs starting from the first stage to the Nth stage is odd. (Each identical box in the
iterative
network has two inputs and two outputs). The optimal logic structure for the box consists
of:
a. One AND gate and one NOR gate
b. One NOR gate and one NAND gate
c. Two XNOR gates
d. One XOR gate
I 1 I 2 I n I n +1 I n + 2
0
Y1 Y2 Yn Yn+1 Yn+2
10. Consider a circuit with N logic nets. If each net can be stuck-at either values 0 and 1, in
how
many ways can the circuit be faulty such that only one net in it can be faulty, and such that
up-to all
nets in it can be faulty?
a. 2 and 2N
b. N and 2^N
c. 2N and 3^N-1
d. 2N and 3N
11. In the circuit shown, all the flip-flops are identical. If the set-up time is 2 ns, clock->Q
delay is 3
ns and hold time is 1 ns, what is the maximum frequency of operation for the circuit?
D1 Q1 D2 Q2 D3 Q3
CLOCK SIGNAL
a. 200 MHz
b. 333 MHz
c. 250 MHz
d. None of the above
12. Which of the following statements is/are true?
I. Combinational circuits may have feedback, sequential circuits do not.
II. Combinational circuits have a memory-less property, sequential circuits do
not.
III. Both combinational and sequential circuits must be controlled by an external
clock.
a. I only
b. II and III only
c. I and II only
d. II only
13. Consider an alternate binary number representation scheme, wherein the number of
ones M, in
a word of N bits, is always the same. This scheme is called the M-out-of-N coding scheme. If
M=N/2, and N=8, what is the efficiency of this coding scheme as against the regular binary
number
representation scheme? (As a hint, consider that the number of unique words represent
able in the
latter representation with N bits is 2^N. Hence the efficiency is 100%)
a. Close to 30%
b. Close to 50%
c. Close to 70%
d. Close to 100%
14. A CPU supports 4 interrupts- I1, I2, I3 and I4. It supports priority of interrupts. Nested
interrupts
are allowed if later interrupt is higher priority than previous one. During a certain period of
time, we
observe the following sequence of entry into and exit from the interrupt service routine:
I1-start---I2-start---I2-end---I4-start---I3-start---I3-end---I4-end---I1-end
From this sequence, what can we infer about the interrupt routines?
a. I3 > I4 > I2 > I1
b. I4 > I3 > I2 > I1
c. I2 > I1; I3 > I4 > I1
d. I2 > I1, I3 > I4 > I2 > I1
15. I decide to build myself a small electric kettle to boil my cup of tea. I need 200 ml of
water for
my cup of tea. Assuming that typical tap water temperature is 25 C and I want the water
boiling in
exactly one minute, then what is the wattage required for the heating element?
[Assume: Boiling point of water is 100 C, 1 Calorie (heat required to change 1 gm of water by
1 C)=
4 joules, 1 ml of water weighs 1 gm.]
a. Data given is insufficient
b. 800 W
c. 300 W
d. 1000 W
e. 250 W
16. The athletics team from REC Trichy is traveling by train. The train slows down, (but does
not
halt) at a small wayside station that has a 100 mts long platform. The sprinter (who can run
100
mts in 10 sec) decides to jump down and get a newspaper and some idlis. He jumps out just
as his
compartment enters the platform and spends 5 secs buying his newspaper that is at the
point
where he jumped out. He then sprints along the platform to buy idlis that is another 50 mts.
He
spends another 5 secs buying the idlis. He is now just 50 mts from the other end of the
platform
where the train is moving out. He begins running in the direction of the train and the only
other
open door in his train is located 50 mts behind the door from where he jumped. At
what(uniform)
speed should the train be traveled if he just misses jumping into the open door at the very
edge of
the platform?
Make the following assumptions
He always runs at his peak speed uniformly
The train travels at uniform speed
He does not wait (other than for the idlis & newspaper) or run baclwards
a. Data given is insufficient
b. 4 m/s
c. 5 m/s
d. 7.5 m/s
e. 10 m/s
1. if a 5-stage pipe-line is flushed and then we have to execute 5 and 12 instructions respectively then no. of cycles will be a. 5 and 12 b. 6 and 13 c. 9 and 16 d.none
2. k-map
ab ---------- c 1 x 0 0 1 x 0 x
solve it
a. A.B B. ~A C. ~B D. A+B
3.CHAR A[10][15] AND INT B[10][15] IS DEFINED WHAT'S THE ADDRESS OF A[3][4] AND B[3][4] IF ADDRESS OD A IS OX1000 AND B IS 0X2000
A. 0X1030 AND 0X20C3 B. OX1031 AND OX20C4 AND SOME OTHERS..
4. int f(int *a) { int b=5; a=&b; }
main() { int i; printf("\n %d",i); f(&i); printf("\n %d",i); }
what's the output .
1.10,5 2,10,10 c.5,5 d. none
5. main() { int i; fork(); fork(); fork(); printf("----"); }
how many times the printf will be executed . a.3 b. 6 c.5 d. 8
6. void f(int i) { int j; for (j=0;j<16;j++) { if (i & (0x8000>>j)) printf("1"); else printf("0"); } } what's the purpose of the program
a. its output is hex representation of i b. bcd c. binary
d. decimal
7.#define f(a,b) a+b #define g(a,b) a*b
main() {
int m; m=2*f(3,g(4,5)); printf("\n m is %d",m); }
what's the value of m a.70 b.50 c.26 d. 69
8. main() { char a[10]; strcpy(a,"\0"); if (a==NULL) printf("\a is null"); else printf("\n a is not null");}
what happens with it . a. compile time error. b. run-time error.
c. a is null d. a is not null.
9. char a[5]="hello"
a. in array we can't do the operation . b. size of a is too large c. size of a is too small
d. nothing wrong with it .
10. local variables can be store by compiler
a. in register or heap
b. in register or stack
c .in stack or heap .
d. global memory.
11. average and worst time complexity in a sorted binary tree is
12. a tree is given and ask to find its meaning (parse-tree)
(expression tree)
ans. ((a+b)-(c*d)) ( not confirmed)
13. convert 40.xxxx into binary .
14. global variable conflicts due to multiple file occurance
is resolved during
a. compile-time
b. run-time
c. link-time
d. load-time
15.
two program is given of factorial.
one with recursion and one without recursion .
question was which program won't run for very big no. input because
of stack overfow .
a. i only (ans.)
b. ii only
c. i& ii both .
c. none
16.
struct a
{
int a;
char b;
int c;
}
union b
{
char a;
int b;
int c;
};
which is correct .
a. size of a is always diff. form size of b.(ans.)
b. size of a is always same form size of b.
c. we can't say anything because of not-homogeneous (not in ordered)
d. size of a can be same if
.THE CURRENT IN 4-OHM RESISTOR IS (CIRCUIT IS GIVEN) ANS:1A
2.WHAT IS THE BELOW CIRCUIT A)MEMORY ELEMENT B)LATCH C)AND GATE D)OSCILLATOR ANS:NOT SURE
3.KARNAUGH MAP ANS:B'(B-BAR)
4.HOW MANY 2:1 MUX ARE REQUIRED TO IMPLEMENT 8:1 MUX ANS:7
5.TWO N_MOS FETS ARE THERE .WHAT IS THE OUTPUT=20 ANS:3VOLTS
6.#DEFINE SUM (A,B) A+B FIND THE VALUE OF X X=3DSUM(2,3)*SUM(3,2) ANS:13(NONE OF THE ABOVE)
7.FIND VALUE OF SIGMA B[i] =B7(-2)POWER I
WHERE B[i] IS THE VALUE AT THE ITH PLACE WHAT IS RESULT IF INPUT IS 0 1 1 0 0 1 1 0 ANS:+34
8.(A EX-OR B)EX-OR B=3D ANS:A
9.AB+CD+EF HOW MANY TWO I/P NAND GATES ARE REQUIRED TO IMPLEMENT THE ABOV= E FUNCTIION
ANS:6
10.HOW MANY BOOLEAN FUNCTIONS WE CAN REALIZE WITH ONE INPUT ANS:4
11. WHAT IS THE TOTAL POWER DISSIPATION IF EACH BULB RATED AT 20 W ,120V= OLTS(IN THE GIVEN FIGURE) ANS:12W
12. TO INCREASE THE SLOPE OF C-MOS .WHAT IS TO BE DONE. ANS:INCREASE W/L RATIO OF P-MOS
13.IF THERE ARE W PEOPLE STANDING IN THE QUEUE AT FIRST IF THE RATE OF RE= MOVAL IS PROPORTIONAL TO REMAINING PERSONS IN THE QUEUE. ANS:DECREASES EXPONENTIALLY
14.SIMLIFY THE FOLLOWING BOOLEAN EXPRESSION=20 A'+A'C+AB'
15.AB'+BC IN THE FOLLOWING ANSWERS WHICH ARE EQUILENT TO GIVEN FORM FUNCT=
IONALLY
ANS:ALL OF THE ABOVE
16.CIRCUIT CONSISTS OF 6 FETS IS GIVEN. O/P IS
ANS:(A+BC)'
17. IN THE FOLLOWING IF 8 I/P 'S IS GIVEN TO AND
GATE .WHICH IS THE FASTE=
ST IMPLEMENTATION=20
ANS: ((( AB)(CD))((EF)(GH)))
18. IF PROBABILITY OF EACH INPUT HIGH IS 0.5
AND LOW IS 0.5.WHAT IS THE PROBABILITY OF O/P AS
HIGH.
CIRCUIT IS GIVEN)OUTPUT Y=3D(AB)+(CD)'
ANS:13/16.
19.OUT PUT OF THE GIVEN CIRCUIT(WHEAT STONE BRIDGE
WITH DIODES)
ANS:0 VOLTS
20.IF C IS THE CAPACITENCE AND V IS THE APPLIED
VOLTAGE AND F IS THE SWIT=
HING FREQUENCY THEN POWER DISSIPATION IS
ANS: F*C*V*V
1. (steeper transition) by:
1. Increasing W/L of PMOS transistor
2. Increasing W/L of NMOS transistor
3. Increasing W/L of both transistors by the same factor
4. Decreasing W/L of both transistor by the same factor
2. Minimum number of 2-input NAND gates that will be required to implement the function: Y = AB + CD + EF is
1. 4
2. 5
3. 6
4. 7
3. Consider a two-level memory hierarchy system M1 & M2. M1 is accessed first and on miss M2 is accessed.
The access of M1 is 2 nanoseconds and the miss penalty (the time to get the data from M2 in case of a miss) is
100 nanoseconds. The probability that a valid data is found in M1 is 0.97. The average memory access time is:
1. 4.94 nanoseconds
2. 3.06 nanoseconds
3 . 5.00 nanoseconds
4.. 5.06 nanoseconds
4. Interrupt latency is the time elapsed between:
1.Occurrence of an interrupt and its detection by the CPU
2. Assertion of an interrupt and the start of the associated ISR
3. Assertion of an interrupt and the completion of the associated ISR
4. Start and completion of associated ISR
5. Which of the following is true for the function (A.B + A’.C + B.C)
1.This function can glitch and can be further reduced
2. This function can neither glitch nor can be further reduced
3. This function can glitch and cannot be further reduced
4. This function cannot glitch but can be further reduced
6. For the two flip-flop configuration below, what is the relationship of the output at B to the clock frequency?
1. Output frequency is 1/4th the clock frequency, with 50% duty cycle
2. Output frequency is 1/3rd the clock frequency, with 50% duty cycle
3. Output frequency is 1/4th the clock frequency, with 25% duty cycle
4. Output frequency is equal to the clock frequency
7. The voltage on Node B is:
1. 0
2. 10
3. –10
8.A CPU supports 250 instructions. Each instruction op-code has these fields:
1. The instruction type (one among 250)
2. A conditional register specification
3. 3 register operands
4. Addressing mode specification for both source operands
9. The CPU has 16 registers and supports 5 addressing modes. What is the instruction op-code length in bits?
32
24
30
36
10. In the iterative network shown, the output Yn of any stage N is 1 if the total number of 1s at the inputs
starting from the first stage to the Nth stage is odd. (Each identical box in the iterative network has two inputs and
two outputs). The optimal logic structure for the box consists of:
1.One AND gate and one NOR gate
2. One NOR gate and one NAND gate
3. Two XNOR gates
4. One XOR gate
11.Consider a circuit with N logic nets. If each net can be stuck-at either values 0 and 1, in how many ways can
the circuit be faulty such that only one net in it can be faulty, and such that up-to all nets in it can be faulty?
1.2 and 2N
2. N and 2^N
3. 2N and 3^N-1
4. 2N and 3N
12. In the circuit shown, all the flip-flops are identical. If the set-up time is 2 ns, clock->Q delay is 3 ns and hold
time is 1 ns, what is the maximum frequency of operation for the circuit?
a. 200 MHz
b. 333 MHz
c. 250 MHz
d. None of the above
13. Which of the following statements is/are true?
I. Combinational circuits may have feedback, sequential circuits do not.
II. Combinational circuits have a ‘memory-less’ property, sequential circuits do not.
III. Both combinational and sequential circuits must be controlled by an external clock.
1. I only
2. II and III only
3. I and II only
4. II only
14.Consider an alternate binary number representation scheme, wherein the number of ones M, in a word of N
bits, is always the same. This scheme is called the M-out-of-N coding scheme. If M=N/2, and N=8, what is the
efficiency of this coding scheme as against the regular binary number representation scheme? (As a hint,
consider that the number of unique words represent able in the latter representation with N bits is 2^N. Hence the
efficiency is 100%)
1.Close to 30%
2. Close to 50%
3. Close to 70%
4. Close to 100%
15.A CPU supports 4 interrupts- I1, I2, I3 and I4. It supports priority of interrupts. Nested interrupts are allowed if
later interrupt is higher priority than previous one. During a certain period of time, we observe the following
sequence of entry into and exit from the interrupt service routine:
1.I1-start---I2-start---I2-end---I4-start---I3-start---I3-end---I4-end---I1-end
From this sequence, what can we infer about the interrupt routines?
1. I3 > I4 > I2 > I1
2. I4 > I3 > I2 > I1
3. I2 > I1; I3 > I4 > I1
4. I2 > I1, I3 > I4 > I2 > I1
16. I decide to build myself a small electric kettle to boil my cup of tea. I need 200 ml of water for my cup of tea.
Assuming that typical tap water temperature is 25 C and I want the water boiling in exactly one minute, then what
is the wattage required for the heating element?
[Assume: Boiling point of water is 100 C, 1 Calorie (heat required to change 1 gm of water by 1 C)= 4 joules, 1 ml
of water weighs 1 gm.]
1. Data given is insufficient
2. 800 W
3. 300 W
4. 1000 W
5. 250 W
17.The athletics team from REC Trichy is traveling by train. The train slows down, (but does not halt) at a small
wayside station that has a 100 mts long platform. The sprinter (who can run 100 mts in 10 sec) decides to jump
down and get a newspaper and some idlis. He jumps out just as his compartment enters the platform and spends
5 secs buying his newspaper that is at the point where he jumped out. He then sprints along the platform to buy
idlis that is another 50 mts. He spends another 5 secs buying the idlis. He is now just 50 mts from the other end
of the platform where the train is moving out. He begins running in the direction of the train and the only other
open door in his train is located 50 mts behind the door from where he jumped. At what(uniform) speed should
the train be traveled if he just misses jumping into the open door at the very edge of the platform?
1.Make the following assumptions
*He always runs at his peak speed uniformly
* The train travels at uniform speed
* He does not wait (other than for the idlis & newspaper) or run baclwards
1. Data given is insufficient
2.4 m/s
3. 5 m/s
4. 7.5 m/s
5. 10 m/s
18.State which of the following gate combinations does not form a universal logic set:
1. 2-input AND + 2-input OR
2. 2-to-1 multiplexer
3. 2-input XOR + inverter
4. 3-input NAND
19.For the circuit shown below, what should the function F be, so that it produces an output of the same
frequency (function F1), and an output of double the frequency (function F2).
1. F1= NOR gate and F2= OR gate
2.F1=NAND gate and F2= AND gate
3. F1=AND gate and F2=XOR gate
4. None of the above
20.The FSM (finite state machine) below starts in state Sa, which is the reset state, and detects a particular
sequence of inputs leading it to state Sc. FSMs have a few characteristics. An autonomous FSM has no inputs.
For a Moore FSM, the output depends on the present state alone. For a Mealy FSM, the output depends on the
present state as well as the inputs. Which of the statements best describes the FSM below?
1. It has two states and is autonomous
2.The information available is insufficient
3. It is a Mealy machine with three states
4. It is a Moor machine with three states
21. In the circuit given below, the switch is opened at time t=0. Voltage across the capacitor at t=infinity is:
1. 2V
2. 3V
3. 5V
4. 7V
22.What is the functionality represented by the following circuit?
1. y= ! (b+ac)
2. y= ! (a+bc)
3. y= ! (a(b+c))
4. y= ! (a+b+c)
23.The value (0xdeadbeef) needs to stored at address 0x400. Which of the below ways will the memory look like
in a big endian machine: 0x403 0x402 0x401 0x400
a. be ef de ad
b. ef be ad de
c. fe eb da ed
d. ed da eb fe
24. In a given CPU-memory sub-system, all accesses to the memory take two cycles. Accesses to memories in
two consecutive cycles can therefore result in incorrect data transfer. Which of the following access mechanisms
guarantees correct data transfer?
1. A read operation followed by a write operation in the next cycle.
2. A write operation followed by a read operation in the next cycle.
3. A NOP between every successive reads & writes
4. None of the above
25.An architecture saves 4 control registers automatically on function entry (and restores them on function
return). Save of each registers costs 1 cycle (so does restore). How many cycles are spent in these tasks (save
and restore) while running the following un-optimized code with n=5:
1.Void fib(int n)
{
if((n==0) || (n==1)) return 1;
return(fib(n-1) + fib(n-2));
}
1. 120
2. 80
3. 125
4. 128
26.The maximum number of unique Boolean functions F(A,B), realizable for a two input (A,B) and single output
(Z) circuit is:
1. 2
2. 6
3. 8