Copy-elision doesn't happen when using move

Question

Why does this example print:

#include <iostream> struct X { X() = default; X(X const&) { std::cout << "copy-constructor\n"; } X(X&&) { std::cout << "move-constructor\n"; } X& operator=(X) { return *this; } }; int main() { X x, y; std::cout << "assign from prvalue calls the "; x = X{}; std::cout << "\nassign from xvalue calls the "; x = std::move(y); }

assign from prvalue calls the
assign from xvalue calls the move-constructor

Both X{} and std::move(y) are rvalues so why does only assigning to X{} cause copy-elision?

implement X& operator=(const X&) and X& operator=(X&&), and you will see — BЈовић
– BЈовић, Commented Mar 9, 2015 at 14:12
@BЈовић But they will be ambiguous with my assignment operator. — template boy
– template boy, Commented Mar 9, 2015 at 14:14

Mike Seymour · Accepted Answer · 2015-03-09 14:24:16Z

Copy elision works in the first case because you're initialising the assignment operator's parameter from a temporary; the temporary can be omitted, constructing the parameter directly instead. In the words of the standard, one of the criteria for elision is:

when a temporary class object that has not been bound to a reference would be copied/moved to a class object with the same cv-unqualified type, the copy/move operation can be omitted by constructing the temporary object directly into the target of the omitted copy/move

In the second case, you're not initialising it from a temporary, but from an existing object. It has already been constructed, in a different location to the target, so the optimisation described above can't be done.

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