Order of member constructor and destructor calls

Solution 1:

In other words, are members guaranteed to be initialized by order of declaration and destroyed in reverse order?

Yes to both. See 12.6.2

6 Initialization shall proceed in the following order:

• First, and only for the constructor of the most derived class as described below, virtual base classes shall be initialized in the order they appear on a depth-first left-to-right traversal of the directed acyclic graph of base classes, where “left-to-right” is the order of appearance of the base class names in the derived class base-specifier-list.

• Then, direct base classes shall be initialized in declaration order as they appear in the base-specifier-list (regardless of the order of the mem-initializers).

• Then, non-static data members shall be initialized in the order they were declared in the class definition (again regardless of the order of the mem-initializers).

• Finally, the compound-statement of the constructor body is executed. [ Note: the declaration order is mandated to ensure that base and member subobjects are destroyed in the reverse order of initialization. —end note ]

Solution 2:

Yes, they are (non-static members that is). See 12.6.2/5 for initialization (construction) and 12.4/6 for destruction.

Solution 3:

Yes, the standard guarantees objects get destructed in the reverse order they were created. The reason is that one object may use another, thus depend on it. Consider:

struct A { };

struct B {
 A &a;
 B(A& a) : a(a) { }
};

int main() {
    A a;
    B b(a);
}

If a were to destruct before b then b would hold an invalid member reference. By destructing the objects in the reverse order in which they were created we guarantee correct destruction.