Pointers

2 A pointer is a variable that points to or references a memory location in which data is stored. Each memory cell in the computer has an address that can be used to access that location so a pointer variable points to a memory location we can access and change the contents of this memory location via the pointer.

3 Pointers Pointer declaration: A pointer is a variable that contains the memory location of another variable. The syntax is as shown below. Start by specifying the type of data stored in the location identified by the pointer. The asterisk tells the compiler that you are creating a pointer variable. Finally you give the name of the variable. type * variable name Example: int *ptr; float *string;

4 Pointers Address operator: Once we declare a pointer variable we must point it to something we can do this by assigning to the pointer the address of the variable you want to point as in the following example: ptr=# This places the address where num is stores into the variable ptr. If num is stored in memory address then the variable ptr has the value

5 Pointers /* A program to illustrate pointer declaration*/ #include main() { int *ptr; int sum; sum=45; ptr=∑ printf("\n Sum is %d",sum); printf("\n The sum pointer is %u", ptr); }

Cox & NgArrays and Pointers6 int i1; int i2; int *ptr1; int *ptr2; i1 = 1; i2 = 2; ptr1 = &i1; ptr2 = ptr1; *ptr1 = 3; i2 = *ptr2; i1: i2: ptr1: 0x1000 0x1004 0x1008 … ptr2: … 0x100C 0x1010 0x x Pointers

7 Pointer expressions & pointer arithmetic: Like other variables pointer variables can be used in expressions. For example if p1 and p2 are properly declared and initialized pointers, then the following statements are valid. y=*p1**p2; sum=sum+*p1; z= 5* - *p2/p1; *p2= *p2 + 10; C allows to add integers to or subtract integers from pointers as well as to subtract one pointer from the other. We can also use short hand operators with the pointers p1+=; sum+=*p2; etc., we can also compare pointers by using relational operators the expressions such as p1 >p2, p1==p2 and p1!=p2 are allowed.

8 Pointers /*Program to illustrate the pointer expression and pointer arithmetic*/ #include main() { int *ptr1,*ptr2; int a,b,x,y,z; a=30;b=6; ptr1=&a; ptr2=&b; x=*ptr1+ *ptr2- 6; y=6*- *ptr1/ *ptr2 +30; printf("\na=%d, b=%d",a,b); printf("\nx=%d,y=%d",x,y); }

9 Pointers Pointers and function: 1. Call by reference 2. Call by value. Call by Reference: When we pass address to a function the parameters receiving the address should be pointers. The process of calling a function by using pointers to pass the address of the variable is known as call by reference. The function which is called by reference can change the values of the variable used in the call.

10 Pointers /* example of call by reference*/ #include void main() { void fncn(int *,int *); int a=20,b=30; printf("\n Value of a and b before function call =%d %d",a,b); fncn(&a,&b); printf("\n Value of a and b after function call =%d %d",a,b); } void fncn(int *p,int *q) { *p=100; *q=200; }

11 Pointers Pointer to arrays: An array is actually very much like pointer. We can declare the arrays first element as a[0] or as int *a because a[0] is an address and *a is also an address the form of declaration is equivalent. The difference is pointer is a variable and can appear on the left of the assignment operator that is lvalue. The array name is constant and cannot appear as the left side of assignment operator.

12 Pointers /* A program to display the contents of array using pointer*/ #include main() { int a[100]; int i,j,n,*ptr; printf("\nEnter the elements of the array\n"); scanf("%d",&n); printf("Enter the array elements"); for(i=0;i< n;i++) { scanf("%d",&a[i]); } printf("Array element are"); for(ptr=a;ptr<(a+n);ptr++) { printf("\nValue of stored at address %u",ptr); }

13 Pointers Multi-Dimensional Arrays int multi[ROWS][COLS]; we can access individual elements of the array multi using either: multi[row][col] or *(*(multi + row) + col) To understand more fully what is going on, let us replace *(multi + row) with X as in: *(X + col) Now, from this we see that X is like a pointer since the expression is de- referenced and we know that col is an integer. The arithmetic being used here is of a special kind called "pointer arithmetic". That means that, since we are talking about an integer array, the address pointed to by (i.e. value of) X + col + 1 must be greater than the address X + col by and amount equal to sizeof(int).

14 Pointers Since we know the memory layout for 2 dimensional arrays, we can determine that in the expression multi + row as used above, multi + row + 1 must increase by value an amount equal to that needed to "point to" the next row, which in this case would be an amount equal to COLS * sizeof(int). That says that if the expression *(*(multi + row) + col) is to be evaluated correctly at run time, the compiler must generate code which takes into consideration the value of COLS, i.e. the 2nd dimension. Because of the equivalence of the two forms of expression, this is true whether we are using the pointer expression as here or the array expression multi[row][col].

15 Pointers Thus, to evaluate either expression, a total of 5 values must be known: The address of the first element of the array, which is returned by the expression multi, i.e., the name of the array. The size of the type of the elements of the array, in this case sizeof(int). The 2nd dimension of the array The specific index value for the first dimension, row in this case. The specific index value for the second dimension, col in this case. Given all of that, consider the problem of designing a function to manipulate the element values of a previously declared array. For example, one which would set all the elements of the array multi to the value 1.

16 Pointers More on Strings Well, we have progressed quite a way in a short time! Let's back up a little and look at what was done in Chapter 3 on copying of strings but in a different light. Consider the following function: char *my_strcpy(char dest[], char source[]) { int i = 0; while (source[i] != '') { dest[i] = source[i]; i++; } dest[i] = ''; return dest; } Recall that strings are arrays of characters. Here we have chosen to use array notation instead of pointer notation to do the actual copying. The results are the same, i.e. the string gets copied using this notation just as accurately as it did before. This raises some interesting points which we will discuss. Since parameters are passed by value, in both the passing of a character pointer or the name of the array as above, what actually gets passed is the address of the first element of each array. Thus, the numerical value of the parameter passed is the same whether we use a character pointer or an array name as a parameter. This would tend to imply that somehow source[i] is the same as *(p+i).

17 Pointers Pointers to Functions Up to this point we have been discussing pointers to data objects. C also permits the declaration of pointers to functions. Pointers to functions have a variety of uses and some of them will be discussed here. Consider the following real problem. You want to write a function that is capable of sorting virtually any collection of data that can be stored in an array. This might be an array of strings, or integers, or floats, or even structures. The sorting algorithm can be the same for all. For example, it could be a simple bubble sort algorithm, or the more complex shell or quick sort algorithm. We'll use a simple bubble sort for demonstration purposes. Sedgewick [1] has described the bubble sort using C code by setting up a function which when passed a pointer to the array would sort it. If we call that function bubble(), a sort program is described by bubble_1.c, which follow

18 Pointers /* Program bubble_3.c from PTRTUT10.HTM 6/13/97 */ #include int arr[10] = { 3,6,1,2,3,8,4,1,7,2}; void bubble(int *p, int N); int compare(int *m, int *n); int main(void) { int i; putchar(' '); for (i = 0; i < 10; i++) { printf("%d ", arr[i]); } bubble(arr,10); putchar(' '); for (i = 0; i < 10; i++) { printf("%d ", arr[i]); } return 0; } void bubble(int *p, int N) { int i, j, t; for (i = N-1; i >= 0; i--) { for (j = 1; j <= i; j++) { if (compare(&p[j-1], &p[j])) { t = p[j-1]; p[j-1] = p[j]; p[j] = t; } int compare(int *m, int *n) { return (*m > *n); }

19 Pointers #include void main(){ int i = 3; int *j; int **k; j=&i; k=&j; printf(" %d ",**k); } Explanation: Memory representation

20 Pointers Here 6024, 8085, 9091 is any arbitrary address, it may be different. Value of k is content of k in memory which is 8085 Value of *k means content of memory location which address k keeps. k keeps address Content of at memory location 8085 is 6024 In the same way **k will equal to 3. Short cut way to calculate: Rule: * and & always cancel to each other i.e. *&a = a So *k = *(&j) since k = &j *&j = j = 6024 And **k = **(&j) = *(*&j) = *j = *(&i) = *&i = i = 3

21 Pointers