此typedef声明有什么作用？

Question

I know what typedef does but this statement seems to be puzzling me.

typedef int (*funcptr)();

This declaration

funcptr pf1,pf2

Means this

int (*pf1)(),(*pf2)();

But then how to use pf1 and pf2 in program. How to take input of these values. And what is the use of it.

Answer 1

Your typedef defines a type for a function pointer. Taking a "value" of a function pointer has no meaning: it is a pointer to a piece of executable code; you need to call it to make it useful.

You call a function through a pointer in the same way as if it were a function known to you by name, ie by appending a parenthesized list of parameters to the pointer.

Here is what you can do with pf1 or pf2 :

// These functions return an int and take no parameters.
// They are compatible with funcptr
int function5() {
    return 5;
}
int functionAsk() {
    int res;
    printf("Enter a value: ");
    scanf("%d", &res);
    return res;
}
// This function does not know what fp1 does, but it can use it
void doSomething(funcptr fp1) {
    int res = fp1();
    printf("Function returned %d", res);
}
// Here is how you can call doSomething with different function pointers
pf1 = functionAsk;
doSomething(pf1);
pf2 = function5;
doSomething(pf2);

Note how in the last four lines two calls of doSomething perform different tasks based on what you pass: the first call prompts the user for an entry, while the second call returns five right away.

Answer 2

This is a very artificial example code, but it illustrates how to use typedef function pointers.

#include <stdio.h>
typedef int (*funcptr)();

int return1(void){return 1;}
int return2(void){return 2;}

int main()
{
    funcptr pf1, pf2;
    pf1 = return1;
    pf2 = return2;
    printf("%d, %d\n", pf1(), pf2());
    return 0;
}

Output: 1, 2

Answer 3

pf1和pf2现在是函数指针...如果希望它们指向函数，则应编写pf1 = yourfunctionname。