Why so many brackets?!

Question

#include <stdio.h>

void * square (const void *num);

int main() {
int x, sq_int;
x = 6;
sq_int = square(&x);
printf("%d squared is %d\n", x, sq_int);

return 0;
}

void* square (const void *num) {
static int result;

result = (*(int *)num) * (*(int *)num);
return(result);
}

This is from the function pointers lesson in C. What the?! Why is there so many brackets at the bottom. This entire thing makes no sense!

Answer 1

I see..

Anyway IMHO the example is not very good, since the current Playground compiler complains the absence of 2 casts, in line
sq_int = square(&x); ...it must be: sq_int = (int)square(&x);

and in line
return(result); ... it must be: return (void *)result;

Furthermore, even adding the two casts - again in current Playground compiler which is 64bit - we must be careful, since we are casting from 64 to 32 bits and viceversa

Answer 2

In fact there is no reason for square() to take "const void *" as argument and return "void *" as result; all this as a collateral effect forces to use confusing casts.

The whole thing can be simplified as:

#include <stdio.h>

int square (int num) {
return num*num;
}

int main() {
int x, sq_int;
x = 6;
sq_int = square(x);
printf("%d squared is %d\n", x, sq_int);

return 0;
}

Answer 3

~ swim ~ Sorry for asking again. I'm having some issues with this code
https://code.sololearn.com/caZnuh2493H2/?ref=app

I've been trying for a while and I just can't get it right. I was hoping I'd get used to C by practicing but it's just winding me up. Also thanks for your answer.