Pointers

Variable Memory Snapshot int nrate = 10; The variable is stored at specific memory address A variable is nothing more than a convenient name for a memory location

Address Operator A variable can be referenced using the address operator & example: scanf(“%f”, &x); This statement specifies that the value read is to be stored at the address of x

Pointer Variables A pointer is a variable that holds the address of a memory location If a variable p holds the address of another variable q, then p is said to point to q If q is a variable at location 100 in memory, then p would have the value 100 (q’s address) 100 200 100 q p

How to declare a pointer variable pointer variables are declared using an asterisk ( * ) The asterisk is called the indirection operator or the de-referencing operator). example: int a, b, *ptr; ptr is a pointer to an integer when a pointer is defined, the type of variable to which it will point must be specified. (i.e. a pointer defined to point to an integer cannot also point to a floating point variable.)

Example int *iPtr; double* dPtr; the variable iPtr is declared to point to an int the variable dPtr is declared to point to a double neither variable has not been initialized in the above example declaring a pointer creates a variable capable of holding an address

Why Pointers? They allow you to refer to large data structures in a compact way They facilitate sharing between different parts of programs They make it possible to get new memory dynamically as your program is running They make it easy to represent relationships between data items

Example a ? iPtr s dPtr int a, *iPtr; char* s; double *dPtr; ? ? ?

More about declaring pointers When using the form int* p, q; the * operator does not distribute. In the above example p is declared to be a pointer to int. q is declared to be an int.

Assigning values to a pointer the assignment operator (=) is defined for pointers the right operand can be any expression that evaluates to the same type as the left the operator & in front of an ordinary variable produces the address of that variable. The operator * in front of a pointer variable returns the contents stored at that address

Example int i=6, j; int *iPtr; iPtr = &i; j = *iPtr; i 6 iPtr j 6

Exercise int a, b; int *c, *d; a = 5; a c = &a; d = &b; *d = 9; b name address memory 1 2 3 4 ? int a, b; int *c, *d; a = 5; c = &a; d = &b; *d = 9; print c, *c, &c print a, b a 5 b 9 c 1 d 2 c=1 *c=5 &c=4 a=5 b=9

Exercise Give a memory snapshot after each set of assignment statements int a=1, b=2, *ptr; ptr = &b; int a=1, b=2, *ptr=&b; a = *ptr;

NULL pointer A pointer can be assigned or compared to the integer zero, or, equivalently, to the symbolic constant NULL, which is defined in <stdio.h>. A pointer variable whose value is NULL is not pointing to anything that can be accessed

Example iPtr s dPtr int *iPtr=0; char *s=0; double *dPtr=NULL;

Pointer Assignments A pointer can point to only one location at a time, but several pointers can point to the same location. /* Declare and initialize variables. */ int x=-5, y = 8, *ptr1, *ptr2; /* Assign both pointers to point to x. */ ptr1 = &x; ptr2 = ptr1; The memory snapshot after these statements are executed is ptr1 ptr2 x y -5 8

Pointer Arithmetic Four arithmetic operations are supported +, -, ++, -- only integers may be used in these operations Arithmetic is performed relative to the variable type being pointed to Example: p++; if p is defined as int *p, p will be incremented by 4 (system dependent) if p is defined as double *p, p will be incremented by 8(system dependent when applied to pointers, ++ means increment pointer to point to next value in memory

Accessing Arrays using Pointers #include <stdio.h> int main() { int a[5]={1,3,5,7,9}; int *ptr; int i=0; ptr = a; while (i<5) { printf("%d\n",*(ptr+i)); i++; }

Comparing Pointers You may compare pointers using relational operators Common comparisons are: check for null pointer (p == NULL) check if two pointers are pointing to the same object (p == q) Is this equivalent to (*p == *q) compare two pointers that are pointing to a common object such as an array.

Call by Value void swap(int a, int b) { int temp; temp = a; a = b; b = temp; return; }

Call by Value int x = 2, y = 3; printf("X = %d Y = %d\n",x,y); swap(x,y); Changes made in function swap are lost when the function execution is over

Call by reference void swap2(int *aptr, int *bptr) { int temp; temp = *aptr; *aptr = *bptr; *bptr = temp; return; }

Call by reference int x = 2, y = 3; int *ptr1, *ptr2; ptr1 = &x; ptr2 = &y; swap2(ptr1,ptr2); printf("X = %d Y = %d\n",*ptr1,*ptr2); x = 2; y = 3; swap2(&x,&y); printf("X = %d Y = %d\n",x,y); Version 1 Version 2

Exercise Write a function to compute the roots of quadratic equation ax^2+bx+c=0. void comproots(int a,int b,int c,double *dptr1, double *dptr2) { *dptr1 = (-b - sqrt(b*b-4*a*c))/(2.0*a); *dptr2 = (-b + sqrt(b*b-4*a*c))/(2.0*a); return; }

Exercise int a,b,c; double root1,root2; printf("Enter Coefficients:\n"); scanf("%d%d%d",&a,&b,&c); computeroots(a,b,c,&root1,&root2); printf("First Root = %lf\n",root1); printf("Second Root = %lf\n",root2);

Exercise Implement a function increaseanddouble(…) which takes 2 integers and increases the value of first by 1 and doubles the value of the second x=3;y=5; … printf(“x=%d y=%d\n“,x,y); increaseanddouble(…); OUTPUT x=3 y=5 x=4 y=10

Dynamic Memory Allocation Dynamically allocated memory is determined at runtime A program may create as many or as few variables as required, offering greater flexibility Dynamic allocation is often used to support data structures such as stacks, queues, linked lists and binary trees. Dynamic memory is finite Dynamically allocated memory may be freed during execution

Dynamic Memory Allocation Memory is allocated using the: malloc function (memory allocation) calloc function (cleared memory allocation) Memory is released using the: free function The size of memory requested by malloc or calloc can be changed using the: realloc function

malloc and calloc Both functions return a pointer to the newly allocated memory If memory can not be allocated, the value returned will be a NULL value The pointer returned by these functions is declared to be a void pointer A cast operator should be used with the returned pointer value to coerce it to the proper pointer type #include <stlib.h> to use these functions

Example of malloc int n = 6, m = 4; double *x; int *p; /* Allocate memory for 6 doubles. */ x = (double *)malloc(n*sizeof(double)); To free the space allocated use free() free(x); X

calloc (allocate memory for arrays) int m = 4; int *p; /* Allocate memory for 4 integers. */ p = (int *)calloc(m,sizeof(int)); p[0]=2; p[1]=3; To free the space allocated use free() free(x); p

Realloc (change size of memory) int m = 4; int *p; /* Allocate memory for 4 integers. */ p = (int *)calloc(m,sizeof(int)); /* change the size of memory block pointed by p */ q = realloc(p,2*m*sizeof(int)); To free the space allocated use free() free(x); p