Order of destruction in the case of multiple inheritance

Question

Is the order of destruction well defined in the case of multiple inheritance?

struct A
{
   ~A(){std::cout << "A\n";}   
};

struct B
{
   ~B(){std::cout << "B\n";}   
};

struct AB : public B, public A
{
    ~AB(){std::cout<<"AB\n";}   
};

int main()
{
    AB ab;
}

For the given code my compiler prints:

AB
B
A

But I use more complex constructs (including CWinApp), I get different results. So is the order well-defined? And if so, what is the ordering rule?

Answer 1

From [class.dtor]:

Bases and members are destroyed in the reverse order of the completion of their constructor (see 12.6.2).

The constructor ordering, from [class.base.init]:

In a non-delegating constructor, initialization proceeds in the following order:
— First, and only for the constructor of the most derived class (1.8), virtual base classes are initialized [ ... ]
— Then, direct base classes are initialized in declaration order as they appear in the base-specifier-list (regardless of the order of the mem-initializers).

For your example:

struct AB : public B, public A

The construction order is B then A then AB. So the destruction order is AB then A thenB.