函数比较C中的整数（指针）

Question

Here's the following function which is supposed to compare the values of two integers a and b and return a positive number if a>b and a negative number otherwise:

 int int_cmp(const void *a, const void *b)
{
    const int *ia = (const int*)a;
    const int *ib = (const int*)b;
    return *ia - *ib;
}

I am not too familiar with constant pointers (or pointers to constant s) and I do not really understand the reasoning behind the function above. I would appreciate it if someone could provide a step-by-step explanation.

Answer 1

suppose, in the caller function, you have two int variables,

int p = 10;
int q = 5;

now , from your main() you are calling int_cmp(&p, &q); to compare their values.

in the receiving function int_cmp() the parameters are made const so that inside the int_cmp() function, the values of int p and int q should not be changed. If the values of a and/or b is changed in the int_cmp() , they will be changed in the main() also, as they have been passed using reference. so, to keep the values unchanged, the const is used.

Next, once the parameters are received in int_cmp() , they are typecasted to int as the arithmetic operators can be safely allowed on pointers of defined variable type.

I hope the atithmatic part is quite straightforward. It is de-referencing the pointers and calculating the difference between the values of the pointers a and b and returning the value of the difference.

Answer 2

I'm guessing this method is used in more general callback that expects a function pointer of the following type

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

This is the only reason I can see to use const void* here instead of const int* .

The reason for const is that comparison should be an operation that reads data only. It shouldn't have any need to mutate the parameters in order to compare them. Hence the standard definition of compare takes const data to encourage implementers to have the correct behavior