c interview questions and answers _ techinterviews
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Answer: 10, 5
10, 5
What will be printed as the result of the operation below:
main()
{
char *ptr = Cisco Systems;
*ptr++; printf(%sn,ptr);
ptr++;
printf(%sn,ptr);
}
Answer:Cisco Systems
isco systems
5.
What will be printed as the result of the operation below:
main()
{
char s1[]=Cisco;
char s2[]= systems;
printf(%s,s1);
}
Answer: Cisco
6.
What will be printed as the result of the operation below:
main(){
char *p1;
char *p2;
p1=(char *)malloc(25);
p2=(char *)malloc(25);
strcpy(p1,Cisco);
strcpy(p2,systems);
strcat(p1,p2);
printf(%s,p1);
}
Answer: Ciscosystems
7.
The following variable is available in file1.c, who can access it?:
static int average;
Answer: all the functions in the file1.c can access the variable.
8.
WHat will be the result of the following code?
#define TRUE 0 // some code
while(TRUE)
{
// some code
}
Answer: This will not go into the loop as TRUE is defined as 0.
9.
What will be printed as the result of the operation below:
int x;
int modifyvalue()
{
return(x+=10);
}
int changevalue(int x)
{
return(x+=1);
}
voidmain()
{
int x=10;x++;
changevalue(x);
x++;
modifyvalue();
printf("First output:%dn",x);
10.
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x++;
changevalue(x);
printf("Second output:%dn",x);
modifyvalue();
printf("Third output:%dn",x);
}
Answer: 12 , 13 , 13
What will be printed as the result of the operation below:
main(){
int x=10, y=15;
x = x++;
y = ++y;
printf(%d %dn,x,y);
}
Answer: 11, 16
11.
What will be printed as the result of the operation below:
main()
{
int a=0;
if(a==0)
printf(Cisco Systemsn);printf(Cisco Systemsn);
}
Answer: Two lines with Cisco Systems will be printed.
12.
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98 COMMENTS ON C INTERVIEW QUESTIONS AND ANSWERS
moimart
Posted 4/7/2006 at 4:16 am |Permalink
#1 Its very very very easy. When printf, p2 points to the 4th position not the
beginning. (from 4th position to 20th position is 0)
Joe Yabyam
Posted 6/14/2006 at 4:46 pm | Permalink
#5 is correct. Both GCC/linux as well as Windows (both VC6 and Cygwin) print as
answered - (Answer:Cisco Systems
isco systems)
This is assuming you #include prior to main().
Joe Yabyam
Posted 6/14/2006 at 4:48 pm | Permalink
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re: #5 is correct That is also if you fix the quotes problem in the strings.
Ankur
Posted 7/4/2006 at 9:12 a m | Permalink
there is some prob with q 5 and q 10
i use turbo c compiler
*ptr++ or *(ptr++)
if i nw print ptr as a charit should be printed as C..
but tc displays error code has no effect..
even with %s it says code has no effect.
with (*ptr)++ it is correct..
! is printed
please ans this query..
also in q no. 10 if i mention x as static
it should persist bw functn calls
but still 12,13,13 is output
tel me why?????????
srikPosted 7/8/2006 at 2:52 pm | Permalink
for #11 it the answer is 10 16.
I got scared for a while there.
sreejesh
Posted 7/21/2006 at 5:28 am | Permalink
can we multiply two long integers by using only integer and string
Scott
Posted 7/21/2006 at 4:20 pm | Permalink
Question 8This fails to compile in gcc:
void test()
{
stvar = 3;
}
static int stvar = 3;
Its a small point, but that seems to be what this test is about.
navya
Posted 8/23/2006 at 3:11 am | Permalink
#10 explanation
1)variable X is intialized to 10
2)x++ will have the value 11
eventhough the function changevalue() have a return stmt there is no variable to store
the value
3) after excecution of the function changevalue() the control comes back to the next
stmt x++ where the value of x becomes 12
4)the modifyvalue() function also dont store the value(similar to the above said
changevalue())
5)so the value of x ie 12 is printed
rest of the excecution is similar please try it out
Divya
Posted 9/13/2006 at 2:53 pm |Permalink
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hi,
what wil be the output of below function return and whats the importance of volatile
here
int sqrt(volatile int *ptr)
{
return (*ptr**ptr);
}
ben lev
Posted 10/16/2006 at 6:17 pm | Permalink
Pretty good questions.
Only one error in your output in question #4
Your output
10, 5
10, 5
should be
10 5
10 5
No comma between the numbers only space.
Thanks
Wei
Posted 10/18/2006 at 2:03 am | Permalink
Explanation to #10:
First line x is global variable, the x in main() is local variable. The fu nction
modifyvalue() only effects global x, but not the local x in main().
As to changevalue(x), its called by value, the x in function changevalue(x) wont effect
the x in main().
bharat biswal
Posted 11/3/2006 at 4:50 am |Permalink
How can you find the number of elements stored in a static array at a time ?
mannu
Posted 11/18/2006 at 4:35 a m | Permalink
hi everybody
main()
{
int x=10, y=15;x = x++;
y = ++y;
printf(%d %d\n,x,y);
}
ans should be
11,15
for x++ x=x+1 i.e 11
y;but due to preincrement value of y after print is 15
can u clarify the ans
Surya
Posted 11/30/2006 at 2:08 pm |Permalink
helloo manu ,
as u said ys value shud be 15 is correct but u should note the line as it is y=++y which
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means the value will be incremented and stored into y itself.
So when it come to next line it will show as 16 only.
mrinu
Posted 12/18/2006 at 2:11 a m | Permalink
hi
++y
andy=++y;
both r one and the same.using the second one that is y=++y is meaningless becau se
++y itself means increment y and store it back in y itself so using y=++y is worth
nothing.
and because there r no expresions used ++y and y++ both give the same answer,
ajay kant singh
Posted 1/20/2007 at 8:49 am |Permalink
In Question 2, at step 1: x=y++ + x++
it 20+35=55and at second statement , ++y + ++x
and 22+56=78
the answrer is :5578
can you explaini it to me..
Anders
Posted 1/22/2007 at 3:00 am |Permalink
Question #3:
This is a peculiarity with printf if I am not mistakensince a separate shift with print
produces the sequence 5 20 5. The printf version is correctly 5 20 1, so if anyone can
explain the printf mangling of the 3rd bit, thank you !
malaram
Posted 1/24/2007 at 1:58 am |Permalink
Hi Maxmelbin
The answer of Q#3 is 5,20,1
because
a>b=a/(square of b)
That means
for a=5a>2=5/(2*2)=5/4=1(take only integer value)
Jessie
Posted 2/16/2007 at 1:04 pm |Permalink
Hi!
I dont understand the 10th question.Can any one of u explain me?
What will be printed as the result of the operation below:
int x;
int modifyvalue(){
return(x+=10);
}
int changevalue(int x)
{
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return(x+=1);
}
void main()
{
int x=10;
x++;
changevalue(x);
x++;
modifyvalue();
printf(First output:%d\n,x);
x++;
changevalue(x);
printf(Second output:%d\n,x);
modifyvalue();
printf(Third output:%d\n,x);
}
Answer: 12 , 13 , 13
sangeeta
Posted 2/16/2007 at 10:59 pm |Permalink
the variable x been defined in main (which is local to main) can be accessed in main,
where as the variable x accessed in the subroutines is the global variable
sangeeta
Posted 2/16/2007 at 11:03 pm | Permalink
in other words x in modifyvalue() is the global variable, in changevalue(x) is the x local
to changevalue, in main() again as it is defined here its local to main.
hope this clears Jessies doubt
Ratheesh
Posted 2/18/2007 at 11:23 pm |Permalink
hi ajay kant singh ,
depends on compiler
Ashish
Posted 3/1/2007 at 12:47 pm |Permalink
Hello All,
Can ne one help me in getting the right answer of the question?
q-> Which one of the following is NOT a default promotion?
Choice 1 If either operand of an arithmetic operator is unsigned long, the other
operand is promoted to unsigned long.
Choice 2 If either operand of an arithmetic operator is unsigned int, the other operand
is promoted to unsigned int.
Choice 3 If either operand of an arithmetic operator is int, the other operand is
promoted to int.
Choice 4 If either operand of an arithmetic operator is double, the other operand is
promoted to double.
Choice 5 If either operand of an arithmetic operator is short, the other operand is
promoted to
sirisha
Posted 3/9/2007 at 11: 12 pm | Permalink
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Reffered to Poly
The answer to the Q # 4 is correct.
The first function is declared as global and hence the x and y values are swapped after
calling the swap(x,y) function. Now the current values of x and y are 10 and 5.
When the second function swap1(x,y) is called the values x=10 and y=5 are passed to
the function. But as this is not call by reference the swapped values are lost once it
comes out of the man function. Hence the original values x=10 and y=5 are printed.
satya
Posted 3/14/2007 at 12:20 am |Permalink
priyanka said,
hi ppl ,
#2. x=20, y=35
now x= y++ + x++;
i.e. 35 + 20 ; // after this x= 21 ,y=36
now y= ++x + ++y ;
i.e. 22 + 37 ;
So the answer sd : 5559(i.e 55 and 59 ,,94 is out of question..Plz explain) .
This answer was wrong..
correct answer is..
x= y++ + x++;
i.e. 35 + 20 ; // after this x= 76 ,y=36
now y= ++x + ++y ;
i.e. 77 + 37 ; // after this x=77 , y=114;
is it clear
Sathish Kumar
Posted 4/18/2007 at 5:02 a m | Permalink
main()
{
int x=20,y=35;
x=y++ + x++;
y= ++y + ++x;
printf(%d%d\n,x,y);
}
Answer : 5794
Can anybody explain me this increments in a order..
Bibin Thomas
Posted 4/19/2007 at 8:00 am | Permalink
Q:2
x=y++ + x++;
y= ++y + ++x;
Q:11
x = x++;
Lots of comments on this!!!
i = i++ is UB(Undefined Behaviour) beacuse youre modifying the same variable more
than once without an intervening sequence point.
For more info read: .
Bibin Thomas
Posted 4/19/2007 at 8:02 am |Permalink
(UB)
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for more info:
http://en.wikipedia.org/wiki/Sequence_point
Me
Posted 6/3/2007 at 2:45 a m | Permalink
16 is incorrect (answer is not 11 16, but 10 16):
$ g++ test2.cpp ;./a10 16
$ g++ version
g++ (GCC) 3.4.4 (cygming special) (gdc 0.12, using dmd 0.125)
Copyright (C) 2004 Free Software Foundation, Inc.
James
Posted 6/7/2007 at 7:23 am |Permalink
Answer to muralipattas Qn,
==============================================================
1.if i hav a declatration like this int a[5]={1,3); why the elements a[2],a[3],a[4] areequal to zero.
2.int a[5];
a[0]=4;
a[1]=5;
then why the values a[2],a[3],a[4] are garbage.
==============================================================
In the first case, as explicit array size is specified, and a shorter initiliazation list is
specified, the unspecified elements are set to zero by the compiler. So you will get
zeroes for the 2,3,4 elements.
In the second case the array has no initialization, so the values are undefined during
compile time. So the array will have garbage values for all locations. And you are
changing the values of indexes 0 and 1 only at runtime.
RameshReddy
Posted 6/15/2007 at 5:44 a m | Permalink
#
satya said,
priyanka said,
hi ppl ,
#2. x=20, y=35
now x= y++ + x++;
i.e. 35 + 20 ; // after this x= 21 ,y=36
now y= ++x + ++y ;
i.e. 22 + 37 ;
So the answer sd : 5559(i.e 55 and 59 ,,94 is out of question..Plz explain) .
This answer was wrong..
correct answer is..
x= y++ + x++;
i.e. 35 + 20 ; // after this x= 76 ,y=36
now y= ++x + ++y ;
i.e. 77 + 37 ; // after this x=77 , y=114;
is it clear
My opinion is like, Answer is compiler dependent.
It behaves different for different compilers.
In GCC its ans is 5693
Its Compiler dependent.
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sapkota
Posted 7/30/2007 at 11:55 a m | Permalink
#
kauai said,
On #1,
while(*p2++ = *p1++); When you do an assignment like this isnt always
true? Hence an infinite loop (or at least until there is a memory access violation) ??
THIS IS TOTALL WRONG ,
since value which is pointed by p2 (ie name) is a string and string always terminat
with 0 so there is no chance at all that causes Access Violation.. Hope it clears
SHYJU KV
Posted 8/13/2007 at 4:00 pm | Permalink
clarification for Question#2
step1:
x=y++ + x++; (note operator precedence ++ have high then +)
so x=21 + 36=57
step2:
y=++y + ++x; (note operator precedence all have same so precedence is right to left)
so y=36 + 58 =94 (++y value assigned only after addition )
Karthik ...
Posted 8/17/2007 at 5:35 am |Permalink
ya ,ans 4 2nd 1 is,
5794
x=20; y=35;
I step: x=y++ + x++;
x=36 + 21 = 57;
II step:y=++y + ++x;
y=36 + 58 = 94; /*here value of x assigns 57 then incremented to 58*/
Nw got it
So ans s 5794
Karthik ...
Posted 8/17/2007 at 5:40 am |Permalink
Can any1 say wat s d o/p of printf(%d);
Hi 2 all , i think d ans 4 d abve qn s
o/p shows some Garbage values ,its my guess,so any other answers / comments
priya
Posted 8/27/2007 at 12:33 pm | Permalink
= y++ + x++;
i.e. 35 + 20 ; // after this x= 76 ,y=36
now y= ++x + ++y ;
i.e. 77 + 37 ; // after this x=77 , y=114;is it clear
explain again
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Biswajit Dash
Posted 9/6/2007 at 1 1:34 am |Permalink
int main()
{
int iX = 40000 , iY = 20000;
if( iX + iY > 0)
printf( I am in if );
else
printf( I am in else );}
Tarak
Posted 11/23/2007 at 4:06 am |Permalink
How can i write a macro for finding squreroot of any number?
Aia
Posted 12/1/2007 at 1:23 pm |Permalink
main()
{
int x=20,y=35;
x=y++ + x++;
y= ++y + ++x;
printf(%d%d\n,x,y);
}
Answer : 5794
UNDEFINED BEHAVIOR
Thats the answer to it.
Aia
Posted 12/1/2007 at 2:12 pm |Permalink
The parts of x = x++ and y = ++y invokes what is called in programming Undefined
Behavior.
When you ch ange the value of a variable more than once in the same sequence point
you are in the land of Undefined Behavior. Meaning you get unpredictable effect.
Depending of compiler implementation or situation. Not a good thing to do.
For further reference read in the C Standards:
ISO Sec. 5.1.2.3, Sec. 6.3, Sec. 6.6, Annex C
Sec. 6.3
and in K&R1 Sec. 2.12 p. 50
K&R2 Sec. 2.12 p. 54
Swapnil
Posted 12/2/2007 at 11:14 am |Permalink
1) what is mean by object?
Ans : Object is a instance of a class.
Mihai Gospodaru
Posted 2/29/2008 at 3:06 pm |Permalink
#5 The answer is displayed correctly, because the string begins with a space Cisco
Systems.
It might be useful to display all pieces of code with Courier New font to avoid creating
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confusion among site users.
DIGPAL
Posted 3/7/2008 at 7:31 a m | Permalink
hi friends,
somebody can tell me what will be the out put of following code and why?
for (i= printf("a");i
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Susanta kumar swain
Posted 11/18/2008 at 1:30 am |Permalink
Class afixi {
Function biswa(){
String s= MY Name Is Biswa .;
}
}
i) How to get the value of s out side of the class ?
ii) How to initialize a class in another class ?iii) How to get the method of a class in another class.
iv) Define the class, object, method ?
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