Why doesn't this C++0x code call the move constructor?
c++, c++11
Solution
Your code has two potential locations for where one may expect the move constructor to get called (but it doesn't):
1) calling std::move 2) during assignment.
Regarding 1), std::move does a simple cast - it does not create an object from a copy - if it did then the move constructor might get invoked by it, but since it does a simple rvalue cast it doesn't get invoked. The definition of std::move is similar to `static_cast<Event&&>(temp)`.
Regarding 2), Initialization and assignment are two entirely different operations (even though some forms of initialization use the '=' symbol). Your code does assignment and therefore uses the default assignment operator which is declared to accept a const lvalue reference. Since you never initialize one event object with another, you won't see your move constructor get invoked. If you declared a move assignment operator: `Event& operator=(Event&& other)`, then your current code would invoke it or if you wrote: `Event a; Event tmp = move(a);` your move constructor, as written, would get invoked.
Problem
For some reason, the following code never calls `Event::Event(Event&& e)` ``` Event a; Event b; Event temp; temp = move(a); a = move(b); b = move(temp); ``` why not? Using `std::swap` calls it once. ``` class Event { public: Event(): myTime(0.0), myNode(NULL) {} Event(fpreal t, Node* n); Event(Event&& other); Event(Event const& other) = delete; ~Event(); bool operator<(Event const& other) const { return myTime < other.myTime; } bool operator>(Event const& other) const { return myTime > other.myTime; } fpreal getTime() const { return myTime; } void setTime(fpreal time) { myTime = time; } Node* getNode() const { return myNode; } private: fpreal myTime; Node* myNode; }; ```