What is the difference between returning auto&& and decltype(auto)?

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I am trying to make template wrapper function, that should forward parameters and return value. And I can't decide what is better to use auto&& or decltype(auto) for return type. I've read Scott Meyers article and understood that it is necessary to return decltype(auto) compared to auto not to strip ref_qualifiers. As far as I understand the same argument works for using auto&& over auto. Now I have following questions:

  1. Am I right, that there is no difference between decltype(auto) and auto&& when we return reference to object?
  2. What happens if we return rvalue object, like: return int{};? Will return value be dangling reference?
  3. What is the difference between decltype(auto) and auto&&? What better fits as forward return type?
decltype-auto forwarding-reference c++14 return-type c++
1 Answers

What is the difference between decltype(auto) and auto&&?

decltype(auto) covers three cases. When returning lvalues, the return type would be T& (lvalue-reference); for xvalues, the return type would be T&& (rvalue-reference); for prvalues, the return type would be T (non-reference, i.e. return by-value).

auto&& covers only two cases. When returning lvalues, the return type would be T& (lvalue-reference); for rvalues, including xvalues and prvalues, the return type would be T&& (rvalue-reference). (Forwarding reference is always a reference.)

What happens if we return rvalue object, like: return int{};? Will return value be dangling reference?

For auto&& the return type is rvalue-reference, so yes, the returned reference is always dangling. For decltype(auto) the return type is non-reference then no such trouble.

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