Why forwarding return value is needed

Question

In the doc of std::forward, it gave the following example:

template<class T>
void wrapper(T&& arg)
{
    foo(forward<decltype(forward<T>(arg).get())>(forward<T>(arg).get()));
}

Why is forwarding of return value needed here? What's the cases where it is different to the following code:

template<class T>
void wrapper(T&& arg)
{
    foo(forward<T>(arg).get());
}

Answer 1

Let's break down the possibilities. T::get could return an lvalue reference (which is an lvalue expression), an rvalue reference (which is an xvalue expression), or a prvalue.

The forward expression will convert the lvalue expression into... an lvalue expression. It will convert the xvalue into... an xvalue. And it will convert a prvalue into an xvalue.

C++'s rules about how arguments bind to parameters in overload resolution are the same for prvalue and xvalue expressions. So the last two will always call the same function.

Therefore, the outer forward accomplishes nothing. Indeed, it is worse than doing nothing at all. Why?

Because prvalues in C++17 and above have guaranteed elision; xvalues do not. If foo takes the parameter by value, the additional forward will manifest an unnecessary temporary, which will then be moved into the argument. If the type is something more complex than an int, then there's a decent chance that you're going to lose some performance.

So don't forward return values which you are going to pass directly as function arguments. If you need to store the value in an intermediary auto&& variable, then you'll need to forward that. But if you're doing it in-situ like this, don't.