pointer operators in c
& ==> address operator
* ==> dereference operator
// Example
int a = 1;
int *ptr; // int * defines a pointer variable pointing at an int
ptr = &a; // the address of 'a' is assigned to the pointer
// 'ptr' is now equal to the address of 'a'
// when dereferenced using *, it returns the value at that address
printf("value of a: %d", *ptr); // prints "value of a: 1"poiner in c
int* pc, c, d;
c = 5;
d = -15;
pc = &c; printf("%d", *pc); // Output: 5
pc = &d; printf("%d", *pc); // Ouptut: -15how to make pointers in c
datatype *var;
variable var actually holds the address of the data(memory where it is stored)
*var lets you access the data stored at that addresspoiner in c
#include <stdio.h>
int main()
{
int var =10;
int *p;
p= &var;
printf ( "Address of var is: %p", &var);
printf ( "\nAddress of var is: %p", p);
printf ( "\nValue of var is: %d", var);
printf ( "\nValue of var is: %d", *p);
printf ( "\nValue of var is: %d", *( &var));
/* Note I have used %p for p's value as it represents an address*/
printf( "\nValue of pointer p is: %p", p);
printf ( "\nAddress of pointer p is: %p", &p);
return 0;
}pointer c
#include <stdio.h>
#include <stdlib.h>
int main()
{
int *ptr, i , n1, n2;
printf("Enter size: ");
scanf("%d", &n1);
ptr = (int*) malloc(n1 * sizeof(int));
printf("Addresses of previously allocated memory: ");
for(i = 0; i < n1; ++i)
printf("%u\n",ptr + i);
printf("\nEnter the new size: ");
scanf("%d", &n2);
// rellocating the memory
ptr = realloc(ptr, n2 * sizeof(int));
printf("Addresses of newly allocated memory: ");
for(i = 0; i < n2; ++i)
printf("%u\n", ptr + i);
free(ptr);
return 0;
}Copyright © 2021 Codeinu
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