dynamic memory allocation in c?

Question

#include <stdio.h>
.
.
.
// Dynamic Array
int **tab=(int**)malloc(sizeof(int*)*n);
for(i=0;i<2;i++)
    {
   tab[i]=(int*)malloc(sizeof(int)*2);
    }
// Array = 0
for(i=0;i<n;i++)
{
    tab[i][0]=0;
    tab[i][1]=0;
}
.
.
.

The program crashes just after the Array = 0 comment. PLease help my assignment is due.

NOTES: n is given by the user and can be any int.

If `n > 2` this will crash when you try to dereference a non-existent pointer in `tab[2]` — Dancrumb, Jan 20 '14 at 22:19
@Dancrumb could you, please, care to elaborate on this? Thanks. — Mechanic45, Jan 20 '14 at 22:20
[Do not cast the result of `malloc`](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc) — pzaenger, Jan 20 '14 at 22:23
You only allocate the first two entries in your `tab` array in your first loop. When you try to access the third one in your second loop, you'll be trying to dereference a NULL pointer, hence a crash. — Dancrumb, Jan 21 '14 at 17:05

sudowoodo · Accepted Answer · 2014-01-20T22:29:11.487

1

It looks like you're only allocating memory for the first 2 pointers of tab and not n elements. Then if n is greater than 2 you're trying to dereference pointers that could point to anything. To fix this just change the 2 in the first for loop to n.

for(i=0; i < n; i++)
{
    tab[i]=(int*)malloc(sizeof(int)*2);
}

edited Jan 20 '14 at 22:29

answered Jan 20 '14 at 22:22

sudowoodo

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dynamic memory allocation in c?

1 Answers1