What does auto&& tell us?

Question

By using auto&& var = <initializer> you are saying: I will accept any initializer regardless of whether it is an lvalue or rvalue expression and I will preserve its constness. This is typically used for forwarding (usually with T&&). The reason this works is because a "universal reference", auto&& or T&&, will bind to anything.

You might say, well why not just use a const auto& because that will also bind to anything? The problem with using a const reference is that it's const! You won't be able to later bind it to any non-const references or invoke any member functions that are not marked const.

As an example, imagine that you want to get a std::vector, take an iterator to its first element and modify the value pointed to by that iterator in some way:

auto&& vec = some_expression_that_may_be_rvalue_or_lvalue;
auto i = std::begin(vec);
(*i)++;

This code will compile just fine regardless of the initializer expression. The alternatives to auto&& fail in the following ways:

auto         => will copy the vector, but we wanted a reference
auto&        => will only bind to modifiable lvalues
const auto&  => will bind to anything but make it const, giving us const_iterator
const auto&& => will bind only to rvalues

So for this, auto&& works perfectly! An example of using auto&& like this is in a range-based for loop. See my other question for more details.

If you then use std::forward on your auto&& reference to preserve the fact that it was originally either an lvalue or an rvalue, your code says: Now that I've got your object from either an lvalue or rvalue expression, I want to preserve whichever valueness it originally had so I can use it most efficiently - this might invalidate it. As in:

auto&& var = some_expression_that_may_be_rvalue_or_lvalue;
// var was initialized with either an lvalue or rvalue, but var itself
// is an lvalue because named rvalues are lvalues
use_it_elsewhere(std::forward<decltype(var)>(var));

This allows use_it_elsewhere to rip its guts out for the sake of performance (avoiding copies) when the original initializer was a modifiable rvalue.

What does this mean as to whether we can or when we can steal resources from var? Well since the auto&& will bind to anything, we cannot possibly try to rip out vars guts ourselves - it may very well be an lvalue or even const. We can however std::forward it to other functions that may totally ravage its insides. As soon as we do this, we should consider var to be in an invalid state.

Now let's apply this to the case of auto&& var = foo();, as given in your question, where foo returns a T by value. In this case we know for sure that the type of var will be deduced as T&&. Since we know for certain that it's an rvalue, we don't need std::forward's permission to steal its resources. In this specific case, knowing that foo returns by value, the reader should just read it as: I'm taking an rvalue reference to the temporary returned from foo, so I can happily move from it.

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As an addendum, I think it's worth mentioning when an expression like some_expression_that_may_be_rvalue_or_lvalue might turn up, other than a "well your code might change" situation. So here's a contrived example:

std::vector<int> global_vec{1, 2, 3, 4};

template <typename T>
T get_vector()
{
  return global_vec;
}

template <typename T>
void foo()
{
  auto&& vec = get_vector<T>();
  auto i = std::begin(vec);
  (*i)++;
  std::cout << vec[0] << std::endl;
}

Here, get_vector<T>() is that lovely expression that could be either an lvalue or rvalue depending on the generic type T. We essentially change the return type of get_vector through the template parameter of foo.

When we call foo<std::vector<int>>, get_vector will return global_vec by value, which gives an rvalue expression. Alternatively, when we call foo<std::vector<int>&>, get_vector will return global_vec by reference, resulting in an lvalue expression.

If we do:

foo<std::vector<int>>();
std::cout << global_vec[0] << std::endl;
foo<std::vector<int>&>();
std::cout << global_vec[0] << std::endl;

We get the following output, as expected:

2
1
2
2

If you were to change the auto&& in the code to any of auto, auto&, const auto&, or const auto&& then we won't get the result we want.

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An alternative way to change program logic based on whether your auto&& reference is initialised with an lvalue or rvalue expression is to use type traits:

if (std::is_lvalue_reference<decltype(var)>::value) {
  // var was initialised with an lvalue expression
} else if (std::is_rvalue_reference<decltype(var)>::value) {
  // var was initialised with an rvalue expression
}

First, I recommend reading this answer of mine as a side-read for a step-by-step explanation on how template argument deduction for universal references works.

Does it mean, we are allowed to steal the resources of var?

Not necessarily. What if foo() all of a sudden returned a reference, or you changed the call but forgot to update the use of var? Or if you're in generic code and the return type of foo() might change depending on your parameters?

Think of auto&& to be exactly the same as the T&& in template<class T> void f(T&& v);, because it's (nearly^†) exactly that. What do you do with universal references in functions, when you need to pass them along or use them in any way? You use std::forward<T>(v) to get the original value category back. If it was an lvalue before being passed to your function, it stays an lvalue after being passed through std::forward. If it was an rvalue, it will become an rvalue again (remember, a named rvalue reference is an lvalue).

So, how do you use var correctly in a generic fashion? Use std::forward<decltype(var)>(var). This will work exactly the same as the std::forward<T>(v) in the function template above. If var is a T&&, you'll get an rvalue back, and if it is T&, you'll get an lvalue back.

So, back on topic: What do auto&& v = f(); and std::forward<decltype(v)>(v) in a codebase tell us? They tell us that v will be acquired and passed on in the most efficient way. Remember, though, that after having forwarded such a variable, it's possible that it's moved-from, so it'd be incorrect use it further without resetting it.

Personally, I use auto&& in generic code when I need a modifyable variable. Perfect-forwarding an rvalue is modifying, since the move operation potentially steals its guts. If I just want to be lazy (i.e., not spell the type name even if I know it) and don't need to modify (e.g., when just printing elements of a range), I'll stick to auto const&.

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† auto is in so far different that auto v = {1,2,3}; will make v an std::initializer_list, whilst f({1,2,3}) will be a deduction failure.

Consider some type T which has a move constructor, and assume

T t( foo() );

uses that move constructor.

Now, let's use an intermediate reference to capture the return from foo:

auto const &ref = foo();

this rules out use of the move constructor, so the return value will have to be copied instead of moved (even if we use std::move here, we can't actually move through a const ref)

T t(std::move(ref));   // invokes T::T(T const&)

However, if we use

auto &&rvref = foo();
// ...
T t(std::move(rvref)); // invokes T::T(T &&)

the move constructor is still available.

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And to address your other questions:

... Are there any reasonable situations when you should use auto&& to tell the reader of your code something ...

The first thing, as Xeo says, is essentially I'm passing X as efficiently as possible, whatever type X is. So, seeing code which uses auto&& internally should communicate that it will use move semantics internally where appropriate.

... like you do when you return a unique_ptr<> to tell that you have exclusive ownership ...

When a function template takes an argument of type T&&, it's saying it may move the object you pass in. Returning unique_ptr explicitly gives ownership to the caller; accepting T&& may remove ownership from the caller (if a move ctor exists, etc.).