括号成员函数的地址错误

问题描述

我发现了一些有趣的东西。错误消息说这一切。在获取非静态成员函数的地址时，不允许使用括号的原因是什么？我在gcc 4.3.4上编译它。

I found something interesting. The error message says it all. What is the reason behind not allowing parentheses while taking the address of a non-static member function? I compiled it on gcc 4.3.4.

#include <iostream>

class myfoo{
    public:
     int foo(int number){
         return (number*10);
     }
};

int main (int argc, char * const argv[]) {

    int (myfoo::*fPtr)(int) = NULL;

    fPtr = &(myfoo::foo);  // main.cpp:14

    return 0;

}

推荐答案

您不能使用带括号的表达式的地址。这表明您重写了

From the error message, it looks like you're not allowed to take the address of a parenthesized expression. It's suggesting that you rewrite

fPtr = &(myfoo::foo);  // main.cpp:14

到

fPtr = &myfoo::foo;

这是由于规范的一部分（&; 5.3.1 / / p>

This is due to a portion of the spec (§5.3.1/3) that reads

我的重点）。我不知道为什么这是一个规则（我实际上不知道这到现在），但这似乎是编译器抱怨。

(my emphasis). I'm not sure why this is a rule (and I didn't actually know this until now), but this seems to be what the compiler is complaining about.

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