Comment by HarHarVeryFunny
14 hours ago
Well, no, because CAN take isn't the same as WILL take.
Changing something to an rvalue means it'll now match a move constructor, but there is no guarantee a move constructor will be used, even if defined, because you've got classes like std::vector that are picky and are explicitly looking for a noexcept move constructor.
In that sense, std::move() is no different than other passing semantics. Just because you wrote at the call site that you want to pass a copy of your object doesn't mean that the callee will actually make a copy of it.
I'm not sure what you are saying.
If we have foo(std::string a, std string b), and then call it like this:
std::string x;
std::string y;
foo(std::move(x), y);
Then x will be moved into a, and y will be copied into b.
The callee has no say in this - it's just the compiler implementing the semantics of the language.
Who says there's only one resolution candidate? A different overload could be defined elsewhere that the compiler prefers for that particular combination of arguments, that doesn't cause a copy. std::move() works the same way. The semantics of the operation is defined not by what's at the call site, but by the context.
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