should std::common_type use std::decay?

问题

Given types A,B, I am concerned with the exact definition of std::common_type<A,B>, disregarding the variadic case std::common_type<A...> for arbitrary types A.... So let

using T = decltype(true ? std::declval<A>() : std::declval<B>());
using C = std::common_type<A,B>;

Now, according to a number of sources, I have found the following relations (skipping typename for brevity):

• cppreference.com: C::type = std::decay<T>::type

• cplusplus.com: C::type = T

• GCC 4.8.1 <type_traits> implementation: C::type = std::decay<T>::type if T is valid, otherwise C does not contain a ::type member ("SFINAE-friendly")

• Clang 3.3 <type_traits> implementation: C::type = std::remove_reference<T>::type

I find the "SFINAE-friendly" version of GCC a minor detail, while std::remove_reference and std::decay practically only differ in built-in arrays and functions, plus cv-qualification, for which again I am not concerned much. So my question is

Should it be decay<T>::type or just T? What is the rationale of using decay<T>::type? Is it only about representing result A() + B() e.g. for arithmetic expressions?

For instance, experimenting a bit, I have found that in the case of the "just T" definition, we have

common_type<int&,int&> = int&
common_type<int&,long&> = long

that is, an lvalue reference is maintained if types are equal. This reflects the fact that

int a, b;
(true ? a : b) = 0;

is valid, while

int a;
long b;
(true ? a : b) = 0;

is not. This semantics of "allowing assignment if types are equal" is exactly what I need in one application, and I tend to believe that common_type and decay should be two independent steps. Should I just use my own definitions?

回答1:

should std::common_type use std::decay?

Yes, see Library Working Group Defect #2141.

Short version (long version, see link above):

• declval<A>() returns a A&&

• common_type is specified via declval, n3337:

  template <class T, class U>
  struct common_type<T, U> {
      typedef decltype(true ? declval<T>() : declval<U>()) type;
  };
  

• common_type<int, int>::type therefore yields int&&, which is unexpected

• proposed resolution is to add decay

  template <class T, class U>
  struct common_type<T, U> {
      typedef decay_t < decltype(true ? declval<T>() : declval<U>()) > type;
  };
  

• common_type<int, int>::type now yields int

来源：https://stackoverflow.com/questions/21975812/should-stdcommon-type-use-stddecay