将函数数组作为函数的参数传递

Question

My C is a bit rusty. Suppose I have the following

char* (*funcs[n])(void);

How would I define or declare another function to take this array as parameter. I tried with

void func(char* (*p)(void));

But gcc complains.

Answer 1

The parameter has the same type as the array, except that you can omit the size:

void print(char* (*p[])(void)) { ... }

You can also use the fact that the array decays into a pointer and declare:

void print(char* (**p)(void)) { ... }

These function signatures are equivalent. In the function, call the function with either of these:

char *a = (*p)();
char *b = (p[0])();

These calls are equivalent, too.

But, as user694733 points out in the comments, everything is made easier by making a typedef for functions of a certain signature, say:

typedef char *(*Charpfunc)(void);

Then you can just use the familiar syntax for your arrays and function parameters:

char *hello(void) { return "hello"; }
char *howdy(void) { return "howdy"; }
char *goodbye(void) { return "goodbye"; }

typedef char *(*Charpfunc)(void);

void print(Charpfunc *p)
{
    while (*p) {
        puts((*p)());
        p++;
    }
}

int main()
{
    Charpfunc funcs[] = {
        hello, howdy, goodbye, NULL
    };

    print(funcs);
    return 0;
}