在C中使用typedef函数指针

Question

I am trying to use function pointer, which is somewhat like this:

#include "stdio.h"

typedef void (*func)(int*, int*);


void func1(int *a, int*b)
{
    printf("Func1\n");
}

void func2(int *a, int*b)
{
    printf("Func2\n");
}

void func3(int *a, int*b)
{
    printf("Func3\n");
}

int main()
{
    int i;
    func f[] = {
      func1, func2, func3
    };
    printf("Hello\n");

    for(i=0; i< 3; i++)
    {
        func fn = f[i];
        *(fn)(&i, &i);
    }
    return 0;
}

I am always getting error: "void value not ignored as it ought to be"

Do not know, how to overcome this. Can anybody please help?

Answer 1

您需要使用如下所示：

(*fn)(&i, &i);

Answer 2

It's either (*fn)(&i, &i) , or just fn(&i,&i) . Otherwise you're trying to dereference the result of the function call!

(Alternatively, five-star-programmers often say (*****fn)(&i, &i) .

Answer 3

This error means you assign the return of a void function to something, which is an error.

Try this:-

(*fn)(&i, &i);

Answer 4

尝试这个：

(*fn)(&i, &i);

Answer 5

Don't write *(fn)(...) - simple fn(...) is enough, no need to dereference anything. What you have means something like "Dereference the return value of fn(...)", but you obviously can't do that.