带有 void* 参数转换的 function 指针

Question

I have no idea on how to look this up, even the title is confusing, even I am confused about what I'm looking for, and the question has for sure already been asked but it's so specific to be found, so here a bit of context:

int comparison(const int* a, const int* b) {
    return *a - *b;
}

int main(int argc, char const *argv[])
{
    int arr[3] = {1,6,-2};
    qsort(arr,3,sizeof(int),comparison);
    return 0;
}

Well, it does work, but the compiler gives me a warning, because qsort wants a function of type:

int(*)(const void*, const void*) 

and comparison is a function of type:

int(*)(const int*, const int*) 

I want to know why the compiler is not happy because it just has to cast the address. It should even be happy to give a type to a void* pointer. Is this really bad? Like an undefined behavior or something? Or just the compiler whining about nothing much?

Answer 1

After other reasons already given, there's another one. Historically there were platforms for which void * and int * had different bit arrangements, I've heard rumor of one where void * and int * were different sizes. That function pointer cast won't always work.

  const int *ia = (const int *)a;
  const int *ib = (const int *)b;

might not compile away to ia = a; but rather to something like ia = a >> 1 ; So there's really got to be a place for those instructions to be.

Answer 2

The qsort function takes as one of its arguments a function of a type you are not passing. So you'll need to change that.

Inside the comparator, you can recast the pointers to the desired type.

Additionally, you need to dereference the values of the const int pointers you are passing into the comparator function:

#include <stdio.h>
#include <stdlib.h>

static int 
comparator(const void *a, const void *b) 
{
    return *(const int *)a - *(const int *)b;
}

static void
printArr(int arr[], int n) 
{ 
    int i; 
    for (i = 0; i < n; ++i) {
        printf("%d ", arr[i]);
    } 
} 

int
main(int argc, const char **argv)
{
    int arr[3] = {1, 6, -2};
    qsort(arr, 3, sizeof(int), comparator);
    printArr(arr, 3);

    return EXIT_SUCCESS;
}

Answer 3

why he (the compiler) is not happy.

qsort() expects a function point of type int (*)(const void *a, const void *b) , not int (*)(const int *a, const int *b) . The compiler could guess its OK and beform a cast, yet it is more productive for the compiler to warn about such problems.

Or just the compiler whining about nothing much?

By warming you, you are allowed to determine the degree of the problem.

---

In addition to @Alex Reynolds good answer, note that *a - *b may overflow, resulting in the wrong comparison.

Instead:

int comparison(const void *a, const void *b) {
  const int *ia = (const int *)a;
  const int *ib = (const int *)b;
  return (*ia > *ab) - (*ia < *ib);
}

Good compilers recognize the (p>q) - (p<q) idiom and emit efficient code.