c++ - 如何区分重载函数的左值和右值成员函数指针？

Question

我知道我可以这样做来区分右值函数名和左值函数指针:

template <typename RET_TYPE, typename...ARGs>
void takeFunction(RET_TYPE(& function)(ARGs...))
{
    cout << "RValue function" << endl;
}

template <typename RET_TYPE, typename...ARGs>
void takeFunction(RET_TYPE(*& function)(ARGs...))
{
    cout << "LValue function" << endl;
}

void function()
{
}

void testFn()
{
    void(*f)() = function;
    takeFunction(function);
    takeFunction(f);
}

而且我希望对成员函数做同样的事情。但是，它似乎没有翻译:

struct S;
void takeMemberFunction(void(S::&function)()) // error C2589: '&' : illegal token on right side of '::'
{
    cout << "RValue member function" << endl;
}

void takeMemberFunction(void(S::*&function)())
{
    cout << "LValue member function" << endl;
}

struct S
{
    void memberFunction()
    {
    }
};

void testMemberFn()
{
    void(S::*mf)() = &S::memberFunction;
    takeMemberFunction(S::memberFunction);
    takeMemberFunction(mf);
}

为什么？

我知道的另一种方法是对常规函数执行此操作:

void takeFunction(void(*&& function)())
{
    cout << "RValue function" << endl;
}

void takeFunction(void(*& function)())
{
    cout << "LValue function" << endl;
}

void function()
{
}

void testFn()
{
    void(*f)() = function;
    takeFunction(&function);
    takeFunction(f);
}

这确实转化为成员函数:

struct S;
void takeMemberFunction(void(S::*&&function)())
{
    cout << "RValue member function" << endl;
}

void takeMemberFunction(void(S::*&function)())
{
    cout << "LValue member function" << endl;
}

struct S
{
    void memberFunction()
    {
    }
};

void testMemberFn()
{
    void(S::*mf)() = &S::memberFunction;
    takeMemberFunction(&S::memberFunction); // error C2664: 'void takeMemberFunction(void (__thiscall S::* &)(void))' : cannot convert argument 1 from 'void (__thiscall S::* )(void)' to 'void (__thiscall S::* &)(void)'
    takeMemberFunction(mf);
}

但我想知道我的第一个例子没有翻译的差异。

Answer 1

我猜这是一个 Visual C++ 错误，如下面的代码(基本上是您问题中的内容)compiles for me在 gcc 和 clang 上，我认为没有理由不期望它:

struct S;

void bar(void (S::*& f)() ) {
    std::cout << "lvalue" << std::endl;
}
void bar(void (S::*&& p)() ) {
    std::cout << "rvalue" << std::endl;
}

struct S {
    void foo() { }  
};

int main() {
    void (S::*f)();

    bar(f);        // prints lvalue
    bar(&S::foo);  // prints rvalue
}

关于您问题的其他部分，请参阅 Why doesn't reference-to-member exist in C++? .