Return memory address of an array using void

Question

When I try to run the program it crashes, but it prints the numbers.Why is that?

void z(int **k)
{
    int i;
    *k=malloc(20*sizeof(int));
    for (i=0;i<20;i++)
        *k[i]=10;

}
int main()
{
    int *k,i;
    z(&k);
    for (i=0;i<20;i++)
    printf("%d\n",k);
    return 0;
}

It happens because you are printing the pointer and not the value where the pointer points to. By the way, free that memory too. — Michi
– Michi, Commented Jan 20, 2016 at 4:38
*k[i]=10; --> (*k)[i]=10;, printf("%d\n",k); --> printf("%d\n",k[i]); — BLUEPIXY
– BLUEPIXY, Commented Jan 20, 2016 at 4:40
@BLUEPIXY Thank you very much.It works now.Why should I place a parenthesis ? — Mitsos
– Mitsos, Commented Jan 20, 2016 at 4:44
It is because of the binding priority. *k[i] meant *(k[i]) — BLUEPIXY
– BLUEPIXY, Commented Jan 20, 2016 at 4:48

Rishikesh Raje · Accepted Answer · 2016-01-20 04:51:07Z

4

The problem is in

*k[i]=10;

By the order of precedence, it should be,

(*k)[i] = 10;

The subscripting operator has a higher precedence than the indirection operator. See Operator Prececence Rules.

Also, in the printf in main, you are printing printf("%d\n",k). This will print the base addresses of the array. If you want the value you should use

printf("%d\n",k[i]);

edited Jan 20, 2016 at 4:51

answered Jan 20, 2016 at 4:43

Rishikesh Raje

8,6222 gold badges18 silver badges33 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Return memory address of an array using void

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related