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#include <vector>
using namespace std;
struct A
{
A(const vector<int>&) {}
A(vector<int>&&) {}
};
A f()
{
vector<int> coll;
return A{ coll }; // Which constructor of A will be called as per C++11?
}
int main()
{
f();
}
Is coll an xvalue in return A{ coll };?
Does C++11 guarantee A(vector<int>&&) will be called when f returns?
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When returning a named local variable or a temporary expression directly, you should avoid the explicit std::move . The compiler must (and will in the future) move automatically in those cases, and adding std::move might affect other optimizations.
commercial-grade C++ compilers won't do that: the return statement will directly construct x itself. Not a copy of x, not a pointer to x, not a reference to x, but x itself.
In particular, std::move produces an xvalue expression that identifies its argument t . It is exactly equivalent to a static_cast to an rvalue reference type.
std::move is actually just a request to move and if the type of the object has not a move constructor/assign-operator defined or generated the move operation will fall back to a copy.
C++11 does not allow coll to be moved from. It only permits implicit moves in return statements when you do return <identifier>, where <identifier> is the name of a local variable. Any expression more complicated than that will not implicitly move.
And expressions more complex than that will not undergo any form of elision.
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