C++11 rvalues and move semantics with return statement

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First example

std::vector<int> return_vector(void)
{
    std::vector<int> tmp {1,2,3,4,5};
    return tmp;
}

std::vector<int> &&rval_ref = return_vector();

The first example returns a temporary which is caught by rval_ref. That temporary will have its life extended beyond the rval_ref definition and you can use it as if you had caught it by value. This is very similar to the following:

const std::vector<int>& rval_ref = return_vector();

except that in my rewrite you obviously can’t use rval_ref in a non-const manner.

Second example

std::vector<int>&& return_vector(void)
{
    std::vector<int> tmp {1,2,3,4,5};
    return std::move(tmp);
}

std::vector<int> &&rval_ref = return_vector();

In the second example you have created a run time error. rval_ref now holds a reference to the destructed tmp inside the function. With any luck, this code would immediately crash.

Third example

std::vector<int> return_vector(void)
{
    std::vector<int> tmp {1,2,3,4,5};
    return std::move(tmp);
}

std::vector<int> &&rval_ref = return_vector();

Your third example is roughly equivalent to your first. The std::move on tmp is unnecessary and can actually be a performance pessimization as it will inhibit return value optimization.

The best way to code what you’re doing is:

Best practice

std::vector<int> return_vector(void)
{
    std::vector<int> tmp {1,2,3,4,5};
    return tmp;
}

std::vector<int> rval_ref = return_vector();

I.e. just as you would in C++03. tmp is implicitly treated as an rvalue in the return statement. It will either be returned via return-value-optimization (no copy, no move), or if the compiler decides it can not perform RVO, then it will use vector’s move constructor to do the return. Only if RVO is not performed, and if the returned type did not have a move constructor would the copy constructor be used for the return.

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