for example:
Beta_ab&& Beta::toAB() const {
return move(Beta_ab(1, 1));
}
Beta_ab&&
Beta::toAB() const {
return move(Beta_ab(1, 1));
}
This returns a dangling reference, just like with the lvalue reference case. After the function returns, the temporary object will get destructed. You should return Beta_ab
by value, like the following
Beta_ab
Beta::toAB() const {
return Beta_ab(1, 1);
}
Now, it's properly moving a temporary Beta_ab
object into the return value of the function. If the compiler can, it will avoid the move altogether, by using RVO (return value optimization). Now, you can do the following
Beta_ab ab = others.toAB();
And it will move construct the temporary into ab
, or do RVO to omit doing a move or copy altogether. I recommend you to read BoostCon09 Rvalue References 101 which explains the matter, and how (N)RVO happens to interact with this.
Your case of returning an rvalue reference would be a good idea in other occasions. Imagine you have a getAB()
function which you often invoke on a temporary. It's not optimal to make it return a const lvalue reference for rvalue temporaries. You may implement it like this
struct Beta {
Beta_ab ab;
Beta_ab const& getAB() const& { return ab; }
Beta_ab && getAB() && { return move(ab); }
};
Note that move
in this case is not optional, because ab
is neither a local automatic nor a temporary rvalue. Now, the ref-qualifier &&
says that the second function is invoked on rvalue temporaries, making the following move, instead of copy
Beta_ab ab = Beta().getAB();