函数指针与函数引用

问题描述

在下面的代码中，函数指针和我认为的函数引用"似乎具有相同的语义:

In the code below, function-pointer and what i considered as "function-reference" seems to have identical semantics:

#include <iostream>
using std::cout;

void func(int a) {
    cout << "Hello" << a << '
';
}
void func2(int a) {
    cout << "Hi" << a << '
';
}

int main() {
    void (& f_ref)(int) = func;
    void (* f_ptr)(int) = func;

    // what i expected to be, and is, correct:
    f_ref(1);
    (*f_ptr)(2);

    // what i expected to be, and is not, wrong:
    (*f_ref)(4); // i even added more stars here like (****f_ref)(4)
    f_ptr(3);    // everything just works!

    // all 4 statements above works just fine

    // the only difference i found, as one would expect:
//  f_ref = func2; // ERROR: read-only reference
    f_ptr = func2; // works fine!
    f_ptr(5);

    return 0;
}

我在 Fedora/Linux 中使用了 gcc 4.7.2 版

I used gcc version 4.7.2 in Fedora/Linux

是否在某些情况下我必须使用一种而不是另一种?

为什么 f_ptr = &func; 有效?既然 func 应该衰减成一个指针?
虽然 f_ptr = &&func; 不起作用(从 void * 隐式转换)

3. Is(Are) there any situation(s) where I must use one over the other?
4. Why f_ptr = &func; works? Since func should be decayed into a pointer?
   While f_ptr = &&func; doesn't work (implicit conversion from void *)

推荐答案

函数和函数引用(即那些类型的 id-expressions)几乎立即衰减为函数指针，因此表达式 func 和 f_ref 在您的情况下实际上成为函数指针.如果你愿意，你也可以调用 (***func)(5) 和 (******f_ref)(6) .

Functions and function references (i.e. id-expressions of those types) decay into function pointers almost immediately, so the expressions func and f_ref actually become function pointers in your case. You can also call (***func)(5) and (******f_ref)(6) if you like.

在您希望 &-operator 像已应用于函数本身一样工作的情况下，最好使用函数引用，例如&func 与 &f_ref 相同，但 &f_ptr 是另一回事.

It may be preferable to use function references in cases where you want the &-operator to work as though it had been applied to the function itself, e.g. &func is the same as &f_ref, but &f_ptr is something else.

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