Thread: Double pointer to int for a dynamic array

Dear C experts,

The below program compiles without any error and prints values as expected.

However, dumps core. I could do this without dumping core on a int *i, however, int **i is giving a core dump.

I would much appreciate if anyone could provide their views on dynamic allocation of such arrays.

The same thing on a char with same sizeof(int) works without a core dump.

Thanks
Nethaji

Code:

#include <stdio.h>
#include <malloc.h>
int main()
{
 int **i, j, k;
 i = calloc(10, sizeof(int));

 for(j = 0; j < 10; j++)
   {
    i[j] = calloc(30, sizeof(int));
   }

 for(j = 0; j < 10; j++)
   {
    for(k = 0; k < 30; k++)
      {
printf("%d == ", k);
       i[j][k] = (k + 10);
printf("%d\n", i[j][k]);
      }
   }

 /*Free memory:*/
 for(j = 0; j < 10; j++)
   {
    free(i[j]);
   }
 free(i);
}

Dear GReaper and Salem,

Thank you both for your views.

Salem, I tried your sizeof(int*) statement and the program does not dump core now. Everything is fine.

Could you please explain what difference it makes in the allocation of memory.

Example:
int *i; is fine with sizeof(int)
while int **i; is fine with sizeof(int*)

The reason is my search to find the answer returned that the difference is sizeof(int) is 4 bytes while sizeof(int*) is 8 bytes.

Why does a simple int needs 8 bytes of memory to store in a double pointer for each of the pointers?

Your time is much appreciated.

Many thanks
Nethaji

Dear GReaper and Salem,

Just an update. The sizeof(int*) is needed only for the allocation call on line number 6, but not for the line 10. However, having sizeof(int*) for line 6 and 10 does not harm the output.

Thank you very much for the explanation Salem.
Best regards
Nethaji