Question

Possible Duplicate:
Allocate memory for 2D array with minimum number of malloc calls

As far my knowledge of concern in C void pointer is automatically converted into appropriated data type. Below is the program in which I found the warning:

initialization from incompatible pointer type [enabled by default]

#define ROW 3
#define COL 2

int main()
{

        void **ptr = malloc( ROW*COL* sizeof(int) );

        int (*p)[COL] = ptr;

        int i, j;

        for( i = 0; i < ROW; ++i )
                for( j = 0; j < COL; ++j )
                        scanf("%d", &p[i][j]);

        for( i = 0; i < ROW; ++i )
        {
                for( j = 0; j < COL; ++j )
                        printf("%d ", p[i][j]);
                printf("\n");
        }
        free(ptr);
        return 0;
}

but it worked perfectly. If void** pointer is type casted into a “pointer to a COL size integer array” then it’s behavior should change and behave like 2D array?

Answer 1

Firstly, void ** pointer is not a void * pointer. void ** and void * are two completely different types. Whatever conversion properties apply to void * don’t apply to your ptr. Your initialization is indeed invalid, as compiler told you in the warning message.

Secondly, there’s absolutely no need for a void ** pointer in your program. Where did it come form anyway? If you wanted to have an “intermediate” pointer ptr in your program, you should have declared it with void * type

void *ptr = malloc(ROW * COL * sizeof(int));
int (*p)[COL] = ptr;

and the warning would disappear. However, as I said in my previous answer, there’s no real need for any intermediate pointer in this case. You can just do

int (*p)[COL] = malloc(ROW * sizeof *p);