What does `auto && e` do in range-based for-loops?

问题

Assuming my current rule when programming with range-based loops says

Use for(auto const &e :...) or for(auto &e:...) when possible over for(auto a: ...).

I base this on my own experience and this question for example.

But after reading about the new terse for loops I wonder, should I not replace my & in my rule with &&? As written here this looks like the Meyers' Universal References.

So, I ask myself, should my new rule either be

Use for(auto const &&e :...) or for(auto &&e:...) when possible ...

or does that not always work and therefore should rather be the quite complicated one

Check if for(auto const &&e :...) or for(auto &&e:...) is possible, then consider for(auto const &e :...) or for(auto &e:...), and only when needed do not use references.

回答1:

When and if you should use auto&& in for loops has been explained very nicely by Howard Hinnant here.

This leaves the question what x in

auto &&x = ...expr...

actually is. And it is handled as if there there were a function template definition

template <class U> void f(const U& u);

and the type of x is deduced by the same tules as u [§7.1.6.4.(7)].

This means it is not handled as a RValue Reference, but as a "Universal/Forwarding Reference" -- the "Reference Collapsing Rules" apply.

This also holds for

const auto &&x = ...expr...

as the example in §7.1.6.4.(7) states, at least for const auto &x.

But, as PiotrS says in the questions comments, any qualifiers nullifies the URef-ness:

no, because neither T in template<class T> void f(const T&&) is a forwarding reference, nor const auto&& is. The fact that T&& occurs in parameter declaration does not imply it is forwarding reference. Only pure T&& with no qualifiers like const or volatile is forwarding reference, meaning it has to be template<class T> void f(T&&) or auto&&, and never const T&& or const auto&&

来源：https://stackoverflow.com/questions/26991393/what-does-auto-e-do-in-range-based-for-loops