Why this program is giving seg fault?

Question

This is my program :

#include<stdio.h>
int main()
{
    int *n;
    int var;
    scanf("%d",n);
    printf("%d",*n);

}

as scanf stores the value at specified address I am giving the address .Then I am trying to print value at address but its giving segfault.

Answer 1

You should allocate memory for pointers like this:

int* n = (int*)malloc(sizeof(int))

Answer 2

It's because a block of memory has not been allocated to contain the integer value referenced by the variable n. You have only initialized a pointer to the memory block, not the memory block itself.

If you instead do the following, the code will work:

#include <stdio.h>
int main()
{
    int n;
    scanf("%d", &n);
    printf("%d", n);
}

Answer 3

var n is pointer, and you didn't malloc memory for it.