In the code below, why doesn't the first call
mkme = mvme_rv
T& operator=(const T&&)
using namespace std;
using T = vector<int>;
T mvme(10, 1), mkme;
T&& mvme_rv = move(mvme); // rvalue ref?
mkme = mvme_rv; // calls T& operator=(const T&)?
cout << mvme.empty(); // 0
mkme = move(mvme_rv); // calls T& operator=(const T&&)?
cout << mvme.empty(); // 1
This line of code:
mkme = mvme_rv;
is a copy and will thus use a copy assignment (
T& operator=(const T&)). The key thing about this is BOTH objects can be used afterwards and should - if implemented correctly - provide two identical objects.
By contrast, this line of code:
mkme = move(mvme_rv);
is a move assignment (
T& operator=(const T&&)). By convention, this will trash the
mvme_rv object (or at least clear it) and make
mkme basically what
mvme_rv was previously.
T&& means a temporary object (aka xvalue) - something that will not last. The
std::move method basically casts the object to a temporary (credit to @richard-Hodges for that wording). This can then be used in the move assignment method.
So finally to answer you question of why doesn't
mkme = mvme_rv dispatch to
T& operator=(const T&&): it's because
mvme_rv isn't a temporary object (aka xavalue).
More about xvalues: http://en.cppreference.com/w/cpp/language/value_category