Why do you use std::move when you have && in C++11?

Question

Possible Duplicate:
What is move semantics?

I recently attended a C++11 seminar and the following tidbit of advice was given.

when you have && and you are unsure, you will almost always use std::move

Could any one explain to me why you should use std::move as opposed to some alternatives and some cases when you should not use std::move?

Answer 1

First, there's probably a misconception in the question I'll address:
Whenever you see T&& t in code (And T is an actual type, not a template type), keep in mind the value category of t is an lvalue(reference), not an rvalue(temporary) anymore. It's very confusing. The T&& merely means that t is constructed from an object that was an rvalue ¹, but t itself is an lvalue, not an rvalue. If it has a name (in this case, t) then it's an lvalue and won't automatically move, but if it has no name (the result of 3+4) then it is an rvalue and will automatically move into it's result if it can. The type (in this case T&&) has almost nothing to do with the value category of the variable (in this case, an lvalue).

That being said, if you have T&& t written in your code, that means you have a reference to a variable that was a temporary, and it is ok to destroy if you want to. If you need to access the variable multiple times, you do not want to std::move from it, or else it would lose it's value. But the last time you acccess t it is safe to std::move it's value to another T if you wish. (And 95% of the time, that's what you want to do). All of this also applies to auto&& variables.

^{1. if T is a template type, T&& is a forwarding reference instead, in which case you use std::forward<T>(t) instead of std::move(t) the last time. See this question.}