std::move on a stack object

Question

I know this is a very basic, probably even embarrassing, question, but I'm having trouble understanding this. If I std::move from something on the stack to another object, can the other object still be used when the original goes out of scope?

#include <iostream>
#include <string>
int
main(int argc, char* argv[])
{
 std::string outer_scope;
 {
 std::string inner_scope = "candy";
 outer_scope = std::move(inner_scope);
 }
 std::cout << outer_scope << std::endl;
 return 0;
}

Is outer_scope still valid where I'm trying to print it?

• Igor Tandetnik: Yes of course. The whole point of move constructor/assignment is to steal contents from temporaries - it would be pretty useless if the object thus constructed could not be used after the temporary in question dies.
• NathanOliver: If you move something then that means what you moved it to is now in control and the thing that had control no longer does.
• MikeMB: Of course it is
• Igor Tandetnik: @πάνταῥεῖ Huh? The code looks pretty straightforward and non-controversial to me. Which rule, in your opinion, does it run afoul of?
• πάντα ῥεῖ: @IgorTandetnik Sorry I misinterpreted for the vice versa action. There's a dupe for that one though.

Answer 1

Yes it's still valid, the innerscope object loses ownership of the content it previously had, and outerscope becomes the owner. std::move is like a vector swap. If you swap outer and inner, destroying inner won't affect the content now owned by outer.