casting a pointer to a float or pointing to a function with a pointer parameter

问题

I am trying to understand exactly what this line of code is, as I am learning C.

int (*f) (float *)

I know the first part is a pointer to an int named f, however the second part is what I am unsure of.

Would this be classed as a pointer to an int named f which is cast to a pointer-to-float

Would this be a pointer named f, pointing to a function with a pointer parameter.

Would be great if someone could help and possible explain why or the difference between the two as I am having a little trouble understanding this concept.

Thanks!

回答1:

That line declares f as a pointer to a function accepting a float * as its argument and returning an int.

The declaration may seem tricky, but it is just following the rules of C. If we wrote

int *f(float *);

Then it would seem as though f is a function returning int *. Thus, the correct precedence must be forced using parenthesis around *f.

回答2:

Here is an example of how f is defined, assigned a function pointer, and then called

#include <stdio.h>

int foo(float *real)
{
    float num = *real + 0.5F;
    return (int)num;
}

int main(void)
{
    float number = 4.9F;
    int (*f) (float *) = foo;           // define f
    printf("%d\n", f(&number));         // calls f
    return 0;
}

Program output:

回答3:

If you had the declaration:

int foo(float *);

it would be clear that it was a function definition. Using (*f) instead of foo gives you a pointer to a function.

回答4:

It declares f as a pointer to a function taking a float * (a float pointer) and returning int. You can then use it like this

float array[5] = {1.0, 2.0, 3.0, 4.0, 5.0};
printf("%d\n", f(array));

来源：https://stackoverflow.com/questions/37197239/casting-a-pointer-to-a-float-or-pointing-to-a-function-with-a-pointer-parameter

标签

pointers

parameters