Question

The following code gives a segmentation fault. I am not able to figure out as to why. Please see..

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int **ptr;
    int *val;
    int x = 7;
    val = &x;
    *ptr = (int *)malloc(10 * sizeof (*val));
    *ptr[0] = *val;
    printf("%d\n", *ptr[0] );

    return 0;
}

on debugging with gdb, it says:

Program received signal SIGSEGV, Segmentation fault.

0x0804843f in main () at temp.c:10

*ptr = (int *)malloc(10 * sizeof (*val));

Any help regarding the matter is appreciated.

Answer 1

int **ptr; 
*ptr = (int *)malloc(10 * sizeof (*val));

First statement declares a double pointer.
Second dereferences the pointer. In order that you are able to dereference it the pointer should point to some valid memory. it does not hence the seg fault.

If you need to allocate enough memory for array of pointers you need:

ptr = malloc(sizeof(int *) * 10); 

Now ptr points to a memory big enough to hold 10 pointers to int.
Each of the array elements which itself is a pointer can now be accessed using ptr[i] where,

i < 10