6. pointers, structs, and arrays...double b; 6. pointers, structs, and arrays einfuhrung¨ in die...

Pointers Dynamic Memory Allocation Structs—n-Tuples Arrays Bubble Sort Strings

6. Pointers, Structs, and Arrays

March 14 & 15, 2011


Einfuhrung in die Programmierung—Introduction to C/C++, Tobias Weinzierl page 1 of 47

http://www5.in.tum.de/wiki/index.php/Dr._rer._nat._Tobias_Weinzierl


Outline

• Recapitulation

• Pointers

• Dynamic Memory Allocation

• Structs

• Arrays

• Bubble Sort

• Strings





Recapitulation

• What is a for loop?

• What is a declaration and what is a definition?

• What is a namespace for?

• What is a header file and how do we use it?

• What is an enum?

• What happens at the end of a scope?

• What is the difference between call-by-value and call-by-reference?

• Why is there a range for all primitive variables in C?





—6.1. Pointers—

Memory

21:22; 20 // int a23; 24; 25; 26;27; 28:

• We have talked a lot about themapping from variables to addresses.However, we have never analysed thereal addresses.

• The memory is enumerated using aninteger.

• Pointers are variables that hold thememory number (address) instead ofa value.

/ / holds a f l o a t i n g po in t/ / number double a ;/ / holds address o f a f l o a t i n g/ / po i n t numberdouble∗ b ;





We Have Already Used Pointers Before

: C/C++ only supports call-by-value. Call-by-reference is not alanguage concept.

vo id foo ( i n t & a ) { / / a c t u a l l y , C/C++ does not suppor t t h i s}

. . .i n t a = 20;foo ( a ) ;

vo id foo ( i n t ∗ a ) { / / address now i s copied ( c a l l−by−value )/ / work on address o f a , not on a d i r e c t l y

}

. . .i n t a = 20;foo ( /∗∗ address o f a ∗ / ) ;





Pointers And Addresses

• The * operator declares a pointer.

i n t a ;i n t ∗p ;

• The & operator returns the address of a variable (address operator or referenceoperator).

p = &a ;

• The * operator makes an operation act on the location an pointers points toinstead of the pointer itself (dereferencing operator.

a += 10;(∗p ) += 10;





Pointers and the Memory

Memory

21:22; 20 // int a23; 24; 22 // pointer p 25; 26;27; 28:

int a;a = 20;

int *p;p = &a;

a++;p++;(*p)++;*p++; // try this at home





Exercise

vo id foo ( i n t & a ) {a+=1;

}

i n t main ( ) {i n t a = 20;foo ( a ) ;s td : : cout << a ;. . .

}

Rewrite exactly this code without a return

statement and without the reference opera-tor.





Solution

vo id foo ( i n t ∗ a ) {(∗a)+=1; / / i n d i r e c t memory access

}

i n t main ( ) {i n t a = 20; / / d i r e c t memory accessfoo (&a ) ;s td : : cout << a ;. . .

}

• There’s three modifications.• Analyse the call-by-value for the pointer.• Which solution is the better/nicer solution?





When to Use Pointers

vo id foo ( i n t ∗ a ) {(∗a )++;. . .a++; / / That could not have happened wi th re ferences

}

• Pointers are difficult to handle (lots of syntactic overhead).• We know how to do it, but do collaborators know?• Avoid it whenever possible. Use C++ references instead.





Pointer Syntax

The pointer operator binds to the right neighbour.

i n t a ;i n t ∗ b1 ;i n t ∗ b2 ;i n t ∗b3 ;i n t ∗c1 , c2 ;i n t ∗c3 , ∗c4 ;i n t ∗ c5 , c6 ;i n t ∗ c7 , ∗c8 ;i n t ∗∗ c9 ;

c7 = c8 ;c7 = ∗c8 ;∗c7 = c8 ;∗c7 = ∗c8 ;

By the way, often people write int* p=0; to make the pointer point to nothing.However, also int* p = 20; would be fine (and for almost 100 percent is a bug).





Pointers and Scopes

i n t ∗p = &a ;f o r ( i n t i =0; i <2; i ++) {

i n t a = 2∗ i ;p = &a ;

s td : : cout << a << std : : endl ;s td : : cout << ∗p << std : : endl

}s td : : cout << a << s td : : endl ; / / does t h i s work?s td : : cout << ∗p << s td : : endl ; / / does t h i s work?

With pointers, we can violate the end-of-scope rules, i.e. we can access variables thatdo not exist anymore. This is the principle of all these buffer overrun malware.





—6.2. Dynamic Memory Allocation—

Switching Off the Lifecycle Management

double ∗p ;

p = new double ; / / now , p po in t s to an e x i s t i n g new v a r i a b l e

∗p = 20;

de le te p ; / / now v a r i a b l e i s f reed , but p s t i l l po in t s/ / to t h i s v a r i a b l e .

∗p = 30;





Two Memory Leaks

double ∗p ;

f o r ( i n t i =0; i <20; i ++) {p = new double ;

}

de le te p ;

f o r ( i n t i =0; i <20; i ++) {double ∗p = new double ;

}

The upper operation creates 20 doubles on the heap, but it destroys only one double inthe end. Consequently, the remaining 19 doubles are lost for the future programexecution. Let’s sketch the memory layout! The second one destroys the pointer but notthe space where the pointer is pointing to. This is why many applications crash afterseveral hours of execution.





—6.3. Structs—n-Tuples—

/∗∗∗ This represents one student i n our on l i ne system∗ /

s t r u c t Student {i n t number ;double averageGrade ;

} ;

• Structs allow us to create user-defined data structures (n-tuples).

• Once declared, we can use them like built-in types.

• C/C++ takes care of the memory layout. This is particularly important for arrays ofstructs.





Declaration and Definition

/ / T e l l s compi ler t h a t something/ / l i k e a StudentID does e x i s ts t r u c t StudentID ;

/∗∗∗ This represents one student∗ i n our on l i ne system∗ /

s t r u c t Student {i n t number ;double averageGrade ;StudentID∗ i d ;

} ;

• For pointers (and references), it is sufficient to tell the compiler that the type doesexist (declaration).

• The declaration is needed for recursive definitions of structs.

• The declaration is needed for more the composition of structs.

We’ll first analyse struct composition.





Struct Composition (Attributes)

/ / T e l l s compi ler t h a t something/ / l i k e a StudentID does e x i s ts t r u c t StudentID {

i n t photoNumber ;bool isWinterTerm ;

} ;

/∗∗∗ This represents one student∗ i n our on l i ne system∗ /

s t r u c t Student {i n t number ;double averageGrade ;StudentID

i d ; / / i t i sn ’ t a p o i n t e r anymore} ;

Take a sheet of paper and sketch the memory layout for an instance of Student if the id

is a pointer and if it is not a pointer.





Memory

21:22; 40 // photoNo23; true // winter term 24; 25; 26; 1234 // number27; 3.4 // average grade28: 22 // pointer

Memory

24; 25; 26; 1234 // number27; 3.4 // average grade28; 40 // photoNo29; true // winter term 30:31:





Structs Referencing Structs

s t r u c t I n tege rEn t r y {i n t value ;I n tege rEn t r y∗ next ;

} ;

Take a sheet of paper and sketch the memory layout for an instance of IntegerEntryreferencing another instance of IntegerEntry. What could be the reasoning behind sucha data structure?





Memory

21:22; 40 // value23; 26 // next 24; 25; 26; 23 // value27; 0 // next28:





Accessing the Elements of a Struct

/ / T e l l s compi ler t h a t something l i k e a StudentID does e x i s ts t r u c t I n tege rEn t r y {

i n t value ;I n tege rEn t r y∗ next ;

} ;

. . .I n tege rEn t r y myEntry ;myEntry . value = 20;myEntry . next = 0 ;

I n tege rEn t r y∗ pEntry = new In tege rEn t r y ;pEntry−>value = 20;pEntry−>next = 0 ;de le te pEntry ;

• The dot gives access to sub-fields.• The -> operator gives access to sub-fields of pointers.• Structs can be used with new as well.





The Single Linked List

/ / T e l l s compi ler t h a t something l i k e a StudentID does e x i s ts t r u c t I n tege rEn t r y {

i n t value ;I n tege rEn t r y∗ next ;

} ;

vo id p r i n t L i s t ( I n tege rEn t r y∗ f i r s t E n t r y ) {whi le ( f i r s t E n t r y !=0 ) {

s td : : cout << f i r s t E n t r y−>value << ” ” ;f i r s t E n t r y = f i r s t E n t r y−>next ;

}}

vo id append ( In tege rEn t r y∗ f i r s t E n t r y , i n t value ) {whi le ( f i r s t E n t r y−>next !=0 ) {

f i r s t E n t r y = f i r s t E n t r y−>next ;}I n tege rEn t r y∗ newEntry = new In tege rEn t r y ;f i r s t E n t r y−>next = newEntry ;newEntry−>next = 0 ;newEntry−>value = value ;

}

Can you rewrite append() recursively?





Remove Element Operation

• Play around with the list and create alist [2, 4, 6, 8].

• Write a remove operation that acceptsan integer k and removes the kth entryfrom the list.

• Invoke it with remove(2).• Print the result to the terminal. It

should return [2, 4, 8].





An Excursus on Lists and Complexity—the big-O notation

f ∈ O(nk ) Doesn’t grow faster than nk .

• Faster or slower refers to the leading term of the runtime polynomial.• Constants or polynomials of lower order are not considered.• Often, one writes = or says is of instead of is contained in.• What does this mean for our list? What is the n there?

• Study the list in terms of n list entries.• If we remove the first element, this is a fixed number of operations (O(1)).• If we append an element, our while loop has to run over all n element before we

can append an element (O(n)).• If we search for an element, we have to study each element exactly once (O(n)).





An Excursus on Lists and Complexity—the big-O notation

f ∈ O(nk ) Doesn’t grow faster than nk .

• Faster or slower refers to the leading term of the runtime polynomial.• Constants or polynomials of lower order are not considered.• Often, one writes = or says is of instead of is contained in.• What does this mean for our list? What is the n there?

• Study the list in terms of n list entries.• If we remove the first element, this is a fixed number of operations (O(1)).• If we append an element, our while loop has to run over all n element before we

can append an element (O(n)).• If we search for an element, we have to study each element exactly once (O(n)).• What could we do to make the append operation run faster?





—6.4. Arrays—Application example: At the university, we wanna keep track of four grades a studentdid throughout his Master’s studies.

double ana lys is1 ;double ana lys is2 ;double l i nea rA lgeb ra ;double s tochas t i c s ;

. . .

/∗∗∗ Takes grades , computes the average , and re tu rns t h i s value .∗ /

double computeAverage ( const double& g1 , const double& g2 , . . . ) {double r e s u l t = g1+g2+g3+g4 ;r e s u l t /= 4 . 0 ;r e t u r n r e s u l t ;

}

Is this impractical, if we wanna store more than four values.





Array Declaration

Memory

21:22; // grade[0]23; // grade[1]24; // grade[2] 25; // grade[3]26;27; 28: 22 // grade (pointer)

double grade [ 4 ] ;

. . .





Array Access

Memory

21:22; // grade[0]23; // grade[1]24; // grade[2] 25; // grade[3]26;27; 28: 22 // grade (pointer)

double grade [ 4 ] ;

. . .

grade [ 0 ] = 1 . 3 ;grade [ 2 ] = 3 . 3 ;grade [ 3 ] = 1 . 0 ;grade [ 1 ] = 4 . 0 ;

Depending on the context, [] either definesthe size of an array (definition) or gives ac-cess to individual entries of an array (ac-cess).





Arrays & Pointers

double grade [ 4 ] ;grade [ 0 ] = 1 . 3 ;grade [ 2 ] = 3 . 3 ;grade [ 3 ] = 1 . 0 ;grade [ 1 ] = 4 . 0 ;

double∗ p = grade ;i f ( grade [0]==∗p ) . . . / / always t ruei f ( grade [1 ]==∗ ( p +1) ) . . . / / always t rue

• An array variable is basically a pointer to the first element of the array.• An array element access internally implies pointer arithmetics and dereferencing.• Again, we thus have to range checks (size of array) at hand at all.





Grades Revisited—We need a range variable

/∗∗∗ Takes grades , computes the average , and re tu rns t h i s value .∗ /

double computeAverage (double∗ grade ,i n t numberOfGrades

) {double r e s u l t = 0 . 0 ;f o r ( i n t i =0; i<numberOfGrades ; i ++) {

. . .}r e s u l t /= 4 . 0 ;r e t u r n r e s u l t ;

}

. . .double grade [ 4 ] ;





Array Arguments

double computeAverage ( double grade [ ] , i n t numberOfGrades ) {. . .

}

double computeAverage ( const double grade [ ] , i n t numberOfGrades ) {. . .

}

This time, the user is not allowed to modify any entry of grade.





Pitfalls

i n t gradesBSc , gradesMSc [ 5 ] ;

gradesMSc [ 5 ] = 2 . 3 ;gradesMSc [ 3 ] + + ;i n t ∗ p0 = gradesMSc ;i n t ∗ p1 = &gradesMSc [ 0 ] ;i n t ∗ p2 = &gradesMSc [ 1 ] ;gradesMSc++; / / t h a t does not workp2++; / / a r r rgh





Dynamic Arrays

double∗ grades = new double [ 4 5 ] ;

de le te [ ] grades ;

• We can create arrays on the heap.

• Size might be a variable, too.

• Corresponding delete has to be a delete[]. delete without brackets just deletesthe first value.

• If we omit delete, we will get a memory leak.

• If we use the array after delete or before new, it points to garbage (remember:grades is only a pointer).





Array As Return Functions

double∗ createThreeRandomGrades ( ) {double r e s u l t [ 3 ] ;r e s u l t [ 0 ] = 1 . 3 ; r e s u l t [ 1 ] = 2 . 7 ; r e s u l t [ 2 ] = 1 . 0 ;r e t u r n r e s u l t ;

}

double∗ createThreeRandomGrades ( ) {double∗ r e s u l t = new double [ 3 ] ;r e s u l t [ 0 ] = 1 . 3 ; r e s u l t [ 1 ] = 2 . 7 ; r e s u l t [ 2 ] = 1 . 0 ;r e t u r n r e s u l t ;

}

• At the end of the scope, the pointer is destroyed always.

• Arrays on the heap are not destroyed.

• This is called a factory mechanism.

• Someone else invoking us has to delete the array.





const Arrays

double∗ createThreeRandomGrades ( ) {double r e s u l t [ 3 ] ;r e s u l t [ 0 ] = 1 . 3 ; r e s u l t [ 1 ] = 2 . 7 ; r e s u l t [ 2 ] = 1 . 0 ;r e t u r n r e s u l t ;

}

double∗ createThreeRandomGrades ( ) {double∗ r e s u l t = new double [ 3 ] ;r e s u l t [ 0 ] = 1 . 3 ; r e s u l t [ 1 ] = 2 . 7 ; r e s u l t [ 2 ] = 1 . 0 ;r e t u r n r e s u l t ;

}

• At the end of the scope, the pointer is destroyed always.

• Arrays on the heap are not destroyed.

• This is called a factory mechanism.

• Someone else invoking us has to delete the array.





Multidimensional Arrays

double mat r i x [ 4 ] [ 4 ] ;

f o r ( i n t i =0; i <4; i ++) {f o r ( i n t j =0; j <4; j ++) {

mat r i x [ i ] [ j ] = i == j ? 1.0 : 0 . 0 ;}

}mat r i x [ 2 ] [ 1 ] = 1 . 0 ;mat r i x [ 1 2 ] = 1 . 0 ;

• What is the semantics of the for loop?

• Multidimensional arrays basically are flat arrays.

• C/C++ uses row-major format (different to Fortran).





Arrays of Structs

s t r u c t Person {i n t age ;double weight ;

} ;

Person couple [ 2 ] ;Person∗ p = couple ;p++;

• C/C++ takes care of the memory padding.

• It stores the entries in the memory in the following order: couple[0].age,couple[0].weight, couple[1].age, couple[1].weight.

• The increment sets the pointer to the subsequent age address.





Outlook C/C++ Arrays vs. Fortran

• Fortran is claimed to be the language for linear algebra as it is faster.

• Fortran does not provide pointers and dynamic data structures.

• Consequently, compiler can keep track of “‘who has access where”.

• Consequently, compiler can optimise aggressivly (it tries to keep book of allpossible values an array could have—side-effect!).

• So, it is all a matter of exclusivity and the const operator.





—6.5. Bubble Sort—

vo id s o r t (i n t e n t r i e s [ ] ,const i n t & numberOfEntr ies

) {bool sor ted = t rue ;wh i le ( sor ted ) {

sor ted = f a l s e ;f o r (

i n t i =0;i<numberOfEntries−1;i ++

) {i f ( e n t r i e s [ i ]>e n t r i e s [ i +1 ] ) {

. . .sor ted = t rue ;

}}

}}





Complexity of Bubble Sort

• Idea of bubble sort: Run over all elements in the list.• Compare two subsequent elements whether they are in the correct order. Swap

them if necessary.• If a swap still had been necessary, run over list again.• How “expensive” is th sorting?

• The number of comparisons is a good metric. So, let n be the number of elementsin our list.

• At most (worst case), we’ll need n runs over the list.• In the average case, we’ll need n/2 runs over the list.• In each run, we have to do n − 1 comparisons.• Overall, the complexity is O(n2).• More intelligent (but more complex) sorting algorithms need only O(n log n)

comparisons.





—6.6. Strings—

• C has no string concept.• C++ has a nice string concept.• C has chars.• And C has arrays.• Thus, C has arrays of chars.• And C has a mapping of numbers to chars.





Codes—The ASCII Code

Code 0 is a reserved value and does not represent a value.





Termination Codes for Arrays

Memory

21:22; 2423; 1224; 71 25; 7126; 1427; 028:29: 30: 22 // s

Lets have the following mapping:

0 Reserved (termination)

12 a

14 o

24 H

71 l





Strings as Arrays

char s [ 6 ] ;s [ 0 ] = ’H ’ ; s [ 1 ] = ’ a ’ ; s [ 2 ] = ’ l ’ ; s [ 3 ] = ’ l ’ ; s [ 4 ] = ’ o ’ ; s [ 5 ] = 0 ;

Lets have the following mapping:

0 Reserved (termination)

12 a

14 o

24 H

71 l





Strings are Arrays

char s [ 6 ] ;s [ 0 ] = ’H ’ ; s [ 1 ] = ’ a ’ ; s [ 2 ] = ’ l ’ ; s [ 3 ] = ’ l ’ ; s [ 4 ] = ’ o ’ ; s [ 5 ] = 0 ;

char∗ myStr ing = ” Ha l lo ” ;

• We don’t want to pass an integer with each string. Thus, C appends a 0 as lastcharacter.

• Thus, the array has one element more than the string has chars.• Pascal follows a different variant.





Operations on Strings

Memory

21:22; 2423; 1224; 71 25; 7126; 1427; 028:29: 30: 22 // s

Write a function length that takes a stringand counts the number of entries. This num-ber then is returned.





Length Operation

i n t leng th ( const char [ ] s ) {i n t leng th = 0;wh i le ( s [ leng th ]==0) {

l eng th ++;}r e t u r n leng th ;

}

char∗ myStr ing = ” Ha l lo ” ;i n t lengthOfMyStr ing = leng th ( myStr ing ) ;

• What is copied here (call-by-value)!• What happens, if we write

char∗ ano therS t r ing = myStr ing ;

• How can we concatenate strings or add additional characters?• How can we copy strings?• How long is the string abc”, ”defg”, and the concatenation of both of them in terms

of a char array.




6. pointers, structs, and arrays...double b; 6. pointers, structs, and arrays einfuhrung¨ in die...

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