What is the behaviour of compiler generated move constructor?

does std::is_move_constructible<T>::value == true implies that T has a usable move constructor?

Either a move constructor or a copy constructor. Remember that the operation of copy construction satisfies all the requirements that are placed upon the operation move construction, and some more.

In Standard terms, a MoveConstructible object is one for which the evaluation of the expression:

T u = rv; 

makes u equivalent to the value of rv before the construction; the state of rv after being moved-from is unspecified. But since it is unspecified, this means the state could even be identical to the one rv had before being moved from: In other words, u could be a copy of rv.

In fact, the Standard defines the CopyConstructible concept to be a refinement of the MoveConstructible concept (so everything which is CopyConstructible is also MoveConstructible, but not vice versa).

if so, what is the default behaviour of it?

The behavior of an implicitly generated move constructor is to perform a member-wise move of the data members of the type for which it is generated.

Per Parahgraph 12.8/15 of the C++11 Standard:

The implicitly-defined copy/move constructor for a non-union class X performs a memberwise copy/move of its bases and members. [ Note: brace-or-equal-initializers of non-static data members are ignored. See also the example in 12.6.2. —end note ]

Moreover:

1 - is f2 move constructed ?

Yes.

2 - if so, shouldn't the rvalue be invalidated?

Moving a pointer is the same as copying it. So no invalidation is going on, neither should it be going on. If you want a move constructor that leaves the moved-from object in a particular state (i.e. sets a pointer data member to nullptr), you have to write your own - or delegate this responsibility to some smart pointer class such as std::unique_ptr.

Notice, that the word "invalidated" is not quite correct here. Move constructors (as well as move assignment operators) are meant to leave the moved-from object in a valid (yet unspecified) state.

In other words, the class invariant needs to be respected - and it should be possible to invoke on a moved-from objects operations that do not have any precondition on its state (usually, destruction and assignment).

does std::is_move_constructible::value == true implies that T has a usable move constructor?

No. It states that you can take an rvalue expression of the object type and construct an object from it. Whether this uses the move constructor or the copy constructor is not relevant to this trait.

is f2 move constructed ?

Yes.

if so, shouldn't the rvalue be invalidated?

No. That's not how movement works.

how can I know if instances of a class can be properly move-constructed(invalidate the old one)?

That is not any definition of "properly move-constructed" that exists. If you want to "invalidate the old one", then you will have to do that yourself.

Move construction generally guarantees nothing about the state of the old object. It will be in a valid but undefined state. Such state very much can be "the same as it was before". Move construction for a pointer is the same as copying the pointer.

If you want to "invalidate" after a move, then you need to write your own move constructor that explicitly does that.

(I'm using VS11)

Then you have no compiler-generated move constructors at all. Not that it would matter, since the move and copy constructors for pointers both do the same thing.