为什么即使在 class 内部获取成员函数指针值也需要 class 名称限定？

Question

当在 class 的成员函数之一中返回指向 class 的成员函数指针时，我仍然必须指定 class。我不能简单地接受地址。例如，this code works fine:

class Foo {
public:
    void func(int param) { cout << param << endl; }
    void (Foo::*getPointer())(int) { return &Foo::func; }
};

但是如果在 getPointer 中我尝试简单地做：return &func 我得到这个错误：

prog.cpp: In member function 'void (Foo::* Foo::getPointer())(int)':
prog.cpp:8:43: error: ISO C++ forbids taking the address of an unqualified or parenthesized non-static member function to form a pointer to member function. Say '&Foo::func' [-fpermissive]
void (Foo::*getPointer())(int) { return &func; }

为什么我必须指定 class，因为这是我所在的上下文？

Answer 1

(...有一个错误的答案...)

在以下上下文中看起来更奇怪：

class Foo {
public:
    virtual void func(int param) { cout << param << endl; }
    void call(int x) { Foo::func(x); }
    void (Foo::*getPointer())(int) { return &Foo::func; }
};

class Bar : public Foo {
public:
    void func(int param) override { cout << "Hello world!" << endl; }
};


int main() {

    Foo *a = new Bar();
    auto p = a->getPointer();
    (a->*p)(4);
    a->call(4);

    return 0;
}

输出为

Hello world
4

调用Foo::func是在Foo中调用funcclass而调用&Foo::func是虚拟调用。

Answer 2

指针和指向成员的指针是不同的类型，我们可以从草案C++标准部分3.9.2 [basic.compound]中看到，其中包括指针的复合类型以及指向非静态 class 成员 和注释的指针：

Static class members are objects or functions, and pointers to them are ordinary pointers to objects or functions

我认为这个问题在 this quote in an answer from Johannes from the Annotated C++ Reference Manual(ARM):

中有很好的描述

Note that the address-of operator must be explicitly used to get a pointer to member; there is no implicit conversion ... Had there been, we would have an ambiguity in the context of a member function ... For example,
void B::f() {
    int B::* p = &B::i;   // ok
    p = B::i;         // error: B::i is an int
    p = &i;           // error: '&i'means '&this->i'
                      // which is an 'int*'

    int *q = &i;      // ok
    q = B::i;         // error: 'B::i is an int
    q = &B::i;        // error: '&B::i' is an 'int B::*'
}

特别是这些行：

int B::* p = &B::i; // OK

和：

p = &i; // error: '&i'means '&this->i' which is an 'int*'

演示合格名称和非合格名称之间的区别。

为什么即使在 class 内部获取成员函数指针值也需要 class 名称限定？

Why does taking a member function pointer value requires class name qualification even from inside of the class?

c++

methods

function-pointers

dereference

member-function-pointers