Reusing a moved container?

The object being in a valid, but undefined state basically means that while the exact state of the object is not guaranteed, it is valid and as such member functions (or non member functions) are guaranteed to work as long as they don't rely on the object having a certain state.

The clear() member function has no preconditions on the state of the object (other than that it is valid, of course) and can therefore be called on moved-from objects. On the other hand for example front() depends on the container being not empty, and can therefore not be called, since it is not guaranteed to be non empty.

Therefore both ver2 and ver3 should both be fine.

From section 17.3.26 of the spec "valid but unspecified state":

an object state that is not specified except that the object’s invariants are met and operations on the object behave as specified for its type [ Example: If an object x of type std::vector<int> is in a valid but unspecified state, x.empty() can be called unconditionally, and x.front() can be called only if x.empty() returns false. —end example ]

Therefore, the object is live. You can perform any operation that does not require a precondition (unless you verify the precondition first).

clear, for example, has no preconditions. And it will return the object to a known state. So just clear it and use it as normal.