Introduction to Pointers in Programming

1. Senem KUMOVA METN CS115 2008-2009 FALL 1 POINTERS && ARRAYS CHAPTER 6

2. Senem KUMOVA METN CS115 2008-2009 FALL 2 POINTERS POINTER is a programming language data type whose value refers directly to (or "points to") another value stored elsewhere in the computer memory using its address

3. Senem KUMOVA METN CS115 2008-2009 FALL 3 POINTERS int x= 5 is a variable and 5 is the value of it printf(value of x : %d, x); &x is the location or address of the x printf(address of x in memory : %d, &x); Pointers are declared in programs and then used to take addresses !!! int x=5; // some stored value int * p; // a pointer that can hold the address of an integer p=&x; x= 5 &x . . .. MEMORY

4. Senem KUMOVA METN CS115 2008-2009 FALL 4 Pointers: & operator 5 5 0x12345678 0x12345678 x x int x = 5 ; int x = 5 ; p p int *p ; int *p ; p = & x ; p = & x ; // adress of x is // stored in p // adress of x is // stored in p 0x12345678 0x12345678 p is a pointer: it contains the address of a piece of data p is a pointer: it contains the address of a piece of data Memory Address Memory Address &x &x

5. Senem KUMOVA METN CS115 2008-2009 FALL 5 POINTERS int * p; // A pointer that can hold an address // of an integer integer pointer int a =0; p= &a; double * q // A pointer that can hold an address // of a double double pointer double b=1.2; q= &b; char * t // A char pointer char c=D; t = &c;

6. Senem KUMOVA METN CS115 2008-2009 FALL 6 POINTERS main() { int * p; // integer pointer int a =0; p= &a; // a equals to *p printf(%d \n, a); printf(%d \n, *p); *p=6; printf(%d \n, a); printf(%d \n, *p); a++; printf(%d \n, a); printf(%d \n, *p); } a=0 p=&a *p

7. Senem KUMOVA METN CS115 2008-2009 FALL 7 POINTERS int a=1, b=2, *p; p=&a; &a &b &p a=1 b=2 ?? will point somewhere in memory &p 1 2 p=&a &a &b

8. Senem KUMOVA METN CS115 2008-2009 FALL 8 POINTERS b=*p; // equivalent to b=a &a &b &p a=1 b=1 ( the value of a ) p=&a THE VALUE POINTED BY A POINTER CAN BE CHANGED BY * name_pointer *p

9. Senem KUMOVA METN CS115 2008-2009 FALL 9 Pointers : EXAMPLE 1 #include void main(void) { int i = 7, j=3 , * k, *m; k = &i; // k ??? *k ??? j=*k; // k ??? *k ??? m=&i; // i ??? *i ??? (*m) ++; // m ??? *m ??? printf( %d %d %d %d, i, j, *k, *m); }

10. Senem KUMOVA METN CS115 2008-2009 FALL 10 Pointers : EXAMPLE 2 #include void main(void) { char a = D; char *p = &a; *p=*p+2; printf(%c %c, a, *p); a=a+1; printf(%c %c, a, *p); }

11. Senem KUMOVA METN CS115 2008-2009 FALL 11 CALL BY VALUE /* Whenever variables are passed as arguments to a function, their values are copied to the function parameters */ void main() { int a=20; incr1 (a); printf(%d: , a); incr2 (a); printf(%d: , a);} void incr1(int x) { x++; } void incr2(int a) { a++; } R E M E M B E R !

12. Senem KUMOVA METN CS115 2008-2009 FALL 12 CALL BY REFERENCE /* Pointers are passed as arguments to a function, their addresses are assigned to the function parameters defined as pointers:*/ void main() { int a=20; incr (&a); printf(%d: , a); } void incr(int * x) { (*x)++; }

13. Senem KUMOVA METN CS115 2008-2009 FALL 13 CALL BY REFERENCE : Example 1 void addi(int *a, int b); void main() { int a=20, int b=30; addi (&a , b) printf(%d %d :, a, b);} void addi(int *a, int b) { int tmp = *a +b; *a= tmp; b= b++;} OUTPUT??

14. Senem KUMOVA METN CS115 2008-2009 FALL 14 void main() { int a=20; int b=30; swap1 (a, b); swap2 (&a, &b); printf(%d %d: , a, b);} void swap1(int x, int y) { int tmp; tmp=x; x=y; y=tmp; } void swap2(int *x, int *y) { int tmp; tmp = *x; // get value pointed by x. *x = *y; // assign value pointed by y to x *y = tmp; } CALL BY REFERENCE Example 2 : SWAP

15. Senem KUMOVA METN CS115 2008-2009 FALL 15 What is an Array? int grade0, grade1, grade2; int grade [3]; Arrays make it possible to represent large number of homogenous values grade0 grade1 grade2 grade[0] grade[1] grade[2]

16. Senem KUMOVA METN CS115 2008-2009 FALL 16 One Dimensional Arrays type name_of_array[size] EXAMPLE int grade[5]; data data grade[1] data grade[2] data data grade[0] grade[3] grade[4] MEMORY first element in array fift ( the last) element in array starts from 0 ends at size - 1

17. Senem KUMOVA METN CS115 2008-2009 FALL 17 One Dimensional Arrays int grade[5], x=9, y; grade[0]= 0; grade[3]= 4; grade[2]= -1; grade[1]=grade[3]+grade[2]; grade[4]= x; x= grade[2]; y= grade[grade[3]]; 0 3 grade[1] -1 grade[2] 4 9 grade[0] grade[3] grade[4] MEMORY x ? y ?

18. Senem KUMOVA METN CS115 2008-2009 FALL 18 INITIALIZATION float f1[]= { 1.0, 1.1, 1.2, 1.3 }; /* declaration without the size */ float f2[4]= { 1.0, 1.1, 1.2, 1.3 }; /* declaration with size */ char z[] =abc; // char arrays char z[] = {a, b, c, \0} int a[80]={0};

19. Senem KUMOVA METN CS115 2008-2009 FALL 19 EXAMPLE1: /* array1.c*/ main() { int x[10], k; for(k=0; k<10; k++) { x[k]=0; printf(x[%d] = %d\n, k, x[k]); //what will be the output ??? } }

20. Senem KUMOVA METN CS115 2008-2009 FALL 20 EXAMPLE2: /* array2.c*/ main() { int x[10], k; for(k=0; k<10; k++) { x[k]=k; printf(x[%d] = %d\n, k, x[k]); //what will be the output ??? } }

21. Senem KUMOVA METN CS115 2008-2009 FALL 21 EXAMPLE3: /* array3.c*/ main() { int x[10], k, sum=0; for(k=0; k<10; k++) scanf(%d, &x[k]); for(k=0; k<10; k++) { printf(%d, x[k]); sum=sum + x[k]; } printf(sum is %d, sum); }

22. Senem KUMOVA METN CS115 2008-2009 FALL 22 Relationship between Pointers and Arrays /*An array name by itself is an adress or a pointer value */ int a[5] , *p; p=&a[0]; // Equals to p=a;

23. Senem KUMOVA METN CS115 2008-2009 FALL 23 Relationship between Pointers and Arrays &a[0] equals to a then a[0] equals to *a &a[1] equals to a+1 then a[1] equals to *(a+1) &a[2] equals to a+2 then a[2] equals to *(a+2) &a[i] equals to a+i then a[i] equals to *(a+i) EXAMPLE: int a [5]={1,2,3,4,5}; int * p; printf(%d,a[0]); printf(%d,*a); printf(%d,a[2]); printf(%d,*(a+2)); p=&a[4]; P=a+4;

24. Senem KUMOVA METN CS115 2008-2009 FALL 24 Relationship between Pointers and Arrays : Example1 main() { int x[10], k; for(k=0; k<10; k++) { x[k]=k; printf(%d\n, x[k]); } } main() { int x[10], k; for(k=0; k<10; k++) { *(x+k)=k; printf(%d\n, *(x+k)); } }

25. Senem KUMOVA METN CS115 2008-2009 FALL 25 Relationship between Pointers and Arrays : Example2 main() { int x[10], k; for(k=0; k<10; k++) { *(x+k)=0; printf(%d\n, *x+k); } } main() { int x[10], k; for(k=0; k<10; k++) { *(x+k)=0; printf(%d\n, *(x+k)); } }

26. Senem KUMOVA METN CS115 2008-2009 FALL 26 Pointer Arithmetic EXAMPLE double a[2],*p,*q; p=a; // p=&a[0] q=p+1; // q=&a[0]+1= &a[1] printf(%d\n, q-p); /* difference in terms of array elements */ printf(%d\n, (int) q-(int) p); /* difference in memory locations */

27. Senem KUMOVA METN CS115 2008-2009 FALL 27 Arrays as Function Arguments main() { int x[9], r; r=sum(x,9); //sum(&x[0],9) } double sum( int a[], int n) { int i; double result =0.0; for (i=0; i

28. Senem KUMOVA METN CS115 2008-2009 FALL 28 Dynamic Memory Allocation Calloc : contiguous memory allocation Malloc : memory allocation

29. Senem KUMOVA METN CS115 2008-2009 FALL 29 CALLOC calloc(n, el_size) an array of n elements, each element having el_size bytes int main() { int *a; //will be used as an array int n; // size of array .... a=calloc(n, sizeof(int)); /* get space for a , and initialize each bit to zero */ free(a); /* each space opened dynamically must be returned */ }

30. Senem KUMOVA METN CS115 2008-2009 FALL 30 MALLOC /* malloc does not initialize memory */ int main() { int *a; //will be used as an array int n; // size of array .... a=malloc(n*sizeof(int)); /* get space for a */ .... free(a); }

31. Senem KUMOVA METN CS115 2008-2009 FALL 31 DYNAMIC ARRAY : Example main() { int * x, k,s; printf(give array size); scanf(%d,&s); x=malloc(s*sizeof(int)); for(k=0; k

32. Senem KUMOVA METN CS115 2008-2009 FALL 32 NEXT WEEK MULTIDIMENSIONAL ARRAYS SORTING ALGORITHMS : BUBBLE SORT MERGE SORT STRINGS && STRING_HANDLING FUNCTIONS ARRAYS OF POINTERS ARGUMENTS TO MAIN