问题:

all. I can't undestand why the bellow code need a cast to work. Someone can explain it?

class Base {
};

class Derived : public Base {
};

class Class {
public:
    Derived member;
};

...

Derived obj;
Base *ptrObj = &obj; // ok, no cast needed

Derived Class::* ptr = &Class::member; // ok
Base    Class::* ptr = &Class::member; // wrong, need cast, why?

回答1:

Because if Base were allowed (covariant), you could then do this, which is a no-no:

Base Class::* ptr = &Class::member;
Class obj;
obj.*ptr = Base();   // <-- assigned a Base into a Derived field?!

At the same time, pointers-to-members cannot be contravariant either, because otherwise you could do this, which is also a no-no:

struct Class2 {
    Base member;
};

Derived Class2::* ptr2 = &Class2::member;
Class2 obj2;
obj2.member = Base();
Derived& d = obj2.*ptr2;  // <-- assigned a Base into a Derived

So, pointers-to-members are neither covariant nor contravariant, but are invariant: the type must match exactly.

回答2:

Ok, I got your point Chris, but your first example works for ordinary pointers. Why should it not work for member pointers too? See the code bellow.

Derived obj;
Base *ptr = &obj;

*ptr = Base(); // it's weird, but ok

The second example will not work even for ordinary pointers, since downcasting is not allowed without cast. So I don't think that should be a explanation.