Why is an assignment to a base class valid, but an assignment to a derived class a compilation error?

Question

This was an interview question. Consider the following:

struct A {}; 
struct B : A {}; 
A a; 
B b; 
a = b;
b = a; 

Why does b = a;

Answer 1

It's true that a B is an A, but an A is not a B, but this fact is only directly applicable when you're working with pointers or references to A's and B's. The problem here is your assignment operator.

struct A {}; 
struct B : A {};

Is equivalent to

struct A {
   A& operator=(const A&);
}; 
struct B : A {
   B& operator=(const B&);
};

So when you're assigning below:

A a; 
B b; 
a = b;

The assignment operator on a can be called with an argument of b, because a B is an A, so b can be passed to the assignment operator as an A&. Note that a's assignment operator only knows about the data that's in an A, and not the stuff in a B, so any members of B that aren't part of A get lost - this is known as 'slicing'.

But when you're trying to assign:

b = a; 

a is of type A, which is not a B, so a can't match the B& parameter to b's assignment operator.

You would think that b=a should just call the inherited A& A::operator=(const A&), but this is not the case. The assignment operator B& B::operator=(const B&) hides the operator that would be inherited from A. It can be restored again with a using A::operator=; declaration.