Move constructor for derived class - c++

Question

I have 2 classes:
template<typename T>
class base{
T t;
public:
base(base &&b): t(std::move(b.t)){}
};
template<typename T, typename T2>
class derived : protected base<T>{
T2 t2;
public:
derived(derived &&d): base<T>(std::move(d)), t2(std::move(d.t2)){}
};
I move entire d object in the derived move-constructor to initialize base part and d becomes invalid but I still need it to use it's part for t2 initialization
Is it possible to do such a thing?

Answer 1

I would say that your construct is correct except for a little syntax error, you need to qualify base<T> in the initializer list:
derived(derived &&d): base<T>(std::move(d)), t2(std::move(d.t2)){}
First, the order of initialization is independant of the order of the initializer list. Draft n4296 says in 12.6.2 Initializing bases and members [class.base.init] § 13
In a non-delegating constructor, initialization proceeds in the following order:
(13.1) — First, and only for the constructor of the most derived class (1.8), virtual base classes are 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 classes in the derived class base-specifier-list.
(13.2) — 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).
(13.3) — Then, non-static data members are initialized in the order they were declared in the class definition
(again regardless of the order of the mem-initializers).
(13.4) — 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 ]
We have also §7 or same chapter that says:
The initialization performed by each mem-initializer constitutes a full-expression. Any
expression in a mem-initializer is evaluated as part of the full-expression that performs the initialization.
My understanding is that standard says that in the move ctor for the derived class, things happens in that order:
move ctor for base class is called
in turn it calls move ctor for T effectively constructing t member of target and eventually zeroing t member of source
move ctor for T2 object is called - at that moment, the end of the full expression has not been reached, and only t member of source has eventually been destroyed
at the end of the full statement, source object is left in an undetermined state and should no longer be used.