C++ 中虚函数指针的比较

问：

假设我想检查一个子类是否实现了它的父级虚拟函数之一（别管这是否闻起来有糟糕的架构......这是一个练习）。如果我想查看两个常规函数是否相同，我可以检查一下。&f == &g

// Plain old functions
void f() {}
void g() {}

...

std::cout << "&f " << &f << "\n";               // "1"  OK, for some reason func ptrs are converted
std::cout << "&g " << &f << "\n";               // "1"  to booleans when printed. I can dig it.
std::cout << "&f == &g " << (&f == &g) << "\n"; // "0"  Good, &f and &g are unequal as expected.

但是对于虚拟成员函数，行为就不同了。

// Base class with a virtual
struct A {
    virtual void f() {}
};

// Subclass which implements f
struct B : public A {
    void f() {}
};

// Subclass which doesn't implement f
struct C : public A {};

...

std::cout << "&A::f " << &A::f << "\n"; // "1"
std::cout << "&B::f " << &B::f << "\n"; // "1"
std::cout << "&C::f " << &C::f << "\n"; // "1" ... okay ... annoying, but alright.

std::cout << "&A::f == &B::f " << (&A::f == &B::f) << "\n"; // "1" DANGER - why does &A::f == &B::f if &f != &g?
std::cout << "&A::f == &C::f " << (&A::f == &C::f) << "\n"; // "1"
std::cout << "&B::f == &C::f " << (&B::f == &C::f) << "\n"; // "1"

std::cout << "(void*)&A::f " << (void*)&A::f << "\n";   // "0x4084b0" Here's what I was actually looking for.
std::cout << "(void*)&B::f " << (void*)&B::f << "\n";   // "0x4084bc" Good - the B::f differs from A::f as it should
std::cout << "(void*)&C::f " << (void*)&C::f << "\n";   // "0x4084b0" Perfect.

std::cout << "(void*)&A::f == (void*)&B::f " << ((void*)&A::f == (void*)&B::f) << "\n"; // "0"
std::cout << "(void*)&A::f == (void*)&C::f " << ((void*)&A::f == (void*)&C::f) << "\n"; // "1"
std::cout << "(void*)&B::f == (void*)&C::f " << ((void*)&B::f == (void*)&C::f) << "\n"; // "0" These are the comparison results I want

所以我的问题在上面的代码中标记。为什么如果？有没有办法在不强制转换的情况下进行我想要的比较（由于 ）会发出嘈杂的编译器警告？DANGER&A::f == &B::f&f != &gvoid*-Wpmf-conversions