When should std::move be used on a function return value?

In the case of return std::move(foo); the move is superfluous because of 12.8/32:

When the criteria for elision of a copy operation are met or would be met save for the fact that the source object is a function parameter, and the object to be copied is designated by an lvalue, overload resolution to select the constructor for the copy is first performed as if the object were designated by an rvalue.

return foo; is a case of NRVO, so copy elision is permitted. foo is an lvalue. So the constructor selected for the "copy" from foo to the return value of meh is required to be the move constructor if one exists.

Adding move does have a potential effect, though: it prevents the move being elided, because return std::move(foo); is not eligible for NRVO.

As far as I know, 12.8/32 lays out the only conditions under which a copy from an lvalue can be replaced by a move. The compiler is not permitted in general to detect that an lvalue is unused after the copy (using DFA, say), and make the change on its own initiative. I'm assuming here that there's an observable difference between the two -- if the observable behavior is the same then the "as-if" rule applies.

So, to answer the question in the title, use std::move on a return value when you want it to be moved and it would not get moved anyway. That is:

• you want it to be moved, and
• it is an lvalue, and
• it is not eligible for copy elision, and
• it is not the name of a by-value function parameter.

Considering that this is quite fiddly and moves are usually cheap, you might like to say that in non-template code you can simplify this a bit. Use std::move when:

• you want it to be moved, and
• it is an lvalue, and
• you can't be bothered worrying about it.

By following the simplified rules you sacrifice some move elision. For types like std::vector that are cheap to move you'll probably never notice (and if you do notice you can optimize). For types like std::array that are expensive to move, or for templates where you have no idea whether moves are cheap or not, you're more likely to be bothered worrying about it.

The move is unnecessary in both cases. In the second case, std::move is superfluous because you are returning a local variable by value, and the compiler will understand that since you're not going to use that local variable anymore, it can be moved from rather than being copied.

On a return value, if the return expression refers directly to the name of a local lvalue (i.e. at this point an xvalue) there is no need for the std::move. On the other hand, if the return expression is not the identifier, it will not be moved automatically, so for example, you would need the explicit std::move in this case:

T foo(bool which) {
   T a = ..., b = ...;
   return std::move(which? a : b);
   // alternatively: return which? std::move(a), std::move(b);
}

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.