C: Why can't I malloc inside the function

Question

I have int *b and a function

void to_Binary(int num, int range, int **bi_res)
{
   int k = num, c = 0, r;
   *bi_res = (int *) calloc(range,sizeof(int));

   while (range >= c) {
        r = k%2; 
        c++; 
        (*bi_res)[range-c] = r; 
        k /= 2;
    }

   return;
}

This is an example of how I send b to the function

 void main
{
   // A small example if you want to replicate. 
   int *b;
   int i = 0;
   to_Binary(56,6,&b);
   while (i < 6) {printf("%d\n",b[i]); i++;}
   return;
} 

but it gives me seg fault in the loop

Note: The function itself works fine and does what it has to do if I access it in other ways but in the case I am required I have to return an array of ints that I can't know the size of

Answer 1

There are at least two problems with the while loop. The first one is that you are using an incorrect expression in this statement

*bi_res[range-c] = r;

You have to write

( *bi_res )[range-c] = r;

The second one is that if you will even use the correct expression nevertheless when range is equal to c you will try to access beyond the allocated array due to this increment within the loop

c++;

Thus the expression

range-c

will have a negative value.

You could write the loop like

while (range != c) {
    //...
}

And this statement

k =/ 2;

contains a syntax error. You mean

k /= 2;

Pay attention to that the return statement is redundant.

And the function main should be declared like

int main( void )

and you should free the allocated memory in main.