c++ operator= weird behviour

Question

Help? I really have no idea what's happening here? Why in line 3 of assignments it calls operator= of A, when its an assignment of B to B?

class A{
public: 
    A& operator=(const A&){cout << "A assignment" << endl;return *this;}
};

class B:public A{
public: 
    A& operator=(const A&){cout << "B assignment" << endl;return *this;}
};

int main() {
    A a;
    B b;
    B b2;       
    a=b; //  output: A assignment
    b=a; //  output: B assignment
    b=b2; //  output: A assignment WHY??
    return 0;
}

possible duplicate of [virtual assignment operator C++](http://stackoverflow.com/questions/669818/virtual-assignment-operator-c) — Cory Kramer, Jun 24 '15 at 19:57
Sorry for the mistake, I was trying to figure it our and changed it. It does compile now. — Michael Papkov, Jun 24 '15 at 20:00
I want to understand why in b=b2 it goes to assignment operator of A, and not B, since I'm assigning to B right? — Michael Papkov, Jun 24 '15 at 20:06
you cant assign `a` to `b`. Error: `binary '=': no operator found which takes a right-hand operand of type 'A' (or there is no acceptable conversion)` — Acha Bill, Jun 24 '15 at 20:08

score 2 · Answer 1 · answered Jun 24 '15 at 20:07

2

There's still a compiler-generated assignment operator in class B (it's overloaded); unlike how this works with constructors, defining one or more assignment operator overloads does not prevent the compiler from generating a copy assignment operator when one is lacking. The compiler generated one calls A::operator=. And it's the better match for an argument of type B.

answered Jun 24 '15 at 20:07

Cheers and hth. - Alf

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How is it a better match tho? They are exactly the same, are they not? – Michael Papkov Jun 24 '15 at 20:09
@MichaelPapkov: one overload (user-defined) takes an `A const&`, the other (compiler-generated) takes a `B const&`. – Cheers and hth. - Alf Jun 24 '15 at 20:10

score 1 · Accepted Answer · answered Jun 24 '15 at 20:09

You've defined an assignment operator in B, but there's also another implicit copy-assignment operator generated by the compiler:

B& B::operator=(B const&);

This is a better match than the one that takes A const& so it is chosen in the assignment b = b2 (since b2 is a B it doesn't require a derived-to-base conversion for the one that takes an A). The implicit copy-assignment operator calls the copy-assignment operator of the base class which you wrote:

B& B::operator=(B const& b) {
    A::operator=(b);
    return *this;
}

This is why it looks like A::operator=(A const&) is being chosen by the assignment.

It seems like you're right, unless I explicitly overload B& B::operator=(B const& b) my self, It does what you say it does. — Michael Papkov, Jun 24 '15 at 20:21

c++ operator= weird behviour

2 Answers2