malloc segmentation fault

hello everybody,
I have written a simple program with malloc ,and I am getting segmentation fault . Can anyone tell me why its so?

Thanks in advance

Code:

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#include<stdio.h>
#include<malloc.h>

int main()
{
        int **a;
        a[0]=(int *)malloc(2*(sizeof(int)));
}

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Quote:

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Originally Posted by BharathKumar

Can anyone tell me why its so?

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You're accessing memory your program does not own.

1) <malloc.h> is a nonstandard header. When using malloc(), you should #include <stdlib.h> instead.

2) You don't have to cast malloc() - see

http://faq.cprogramming.com/cgi-bin/...&id=1043284351

3) You need to allocate space for a[0], a[1], etc. BEFORE using malloc() to assign memory to each of them. For example, if you wanted to allocate space for a[0], a[1], and a[2], you could write

Code:

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  int **a;
  a = malloc(3*sizeof(int *)); /* allocates space for a[0], a[1], a[2] */
  a[0] = malloc(2*sizeof(int)); /* allocates space for a[0][0], a[0][1] */

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4) Of course you should check for NULL after calling each malloc() - I neglected to do this in the example above.

Thank You Robatino .

And you should free the memory you've allocated, once your done with it :)

Eg:

Code:

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#include <stdlib.h>

int main(void)
{
    const size_t rows = 3, cols = 5;
    int ** a;
    size_t c;
   
    /* allocate rows */
    if((a = malloc(rows * sizeof(int *))) == NULL)
    {
        perror("Failed to allocate rows");
        return 1;
    }

    /* allocate cols */
    for(c = 0; c < rows; c++)
    {
        if((a[c] = malloc(cols * sizeof(int))) == NULL)
        {
            perror("Failed to allocate cols");
            return 2;  /* ideally you should free the cols and rows, but cbf :) */
        }
    }

    /* do whatever with rows and cols */

    /* free cols */
    for(c = 0; c < rows; c++)
    {
        free(a[c]);
        a[c] = NULL;
    }

    /* free rows */
    free(a);
    a = NULL;
    return 0;
}

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Thank You Zacs7 ,as I had missed that part.