Making std::underlying_type SFINAE-friendly

Document number: P0340R0
Date: 2016-05-30
Audience: Library Evolution Working Group
Reply to: R. "Tim" Song <rs2740@gmail.com>

Abstract

This paper proposes making std::underlying_type SFINAE-friendly. In particular, it would make instantiating std::underlying_type<T> for a non-enumeration type T well-defined and result in an empty struct.

Motivation

Currently, std::underlying_type<T> requires T to be a complete enumeration type. instantiating it with any other type results in undefined behavior, typically a hard error. This makes it tricky to use in SFINAE contexts. For example, if one wants to write a function template that want to constrain a template argument to "enumeration type whose underlying type is int", the "obvious" approach would be something along the lines of

template<class T>
std::enable_if_t<std::is_enum_v<T> &&
                 std::is_same_v<std::underlying_type_t<T>, int>> foo(T t);

Unfortunately, this won't work; writing foo(0) will almost certainly result in a hard error, even if there is a void foo(int); overload available. Instead, actual evaluation of std::underlying_type<T> must be deferred until T is known to be an enum, with something like

template<class T>
std::enable_if_t<std::is_same_v<typename std::enable_if_t<std::is_enum_v<T>, std::underlying_type<T>>::type, int>> foo(T t);

Not only is this harder to write (typename ...::type), it is also harder to understand (nested enable_ifs).

This is a recurring problem on StackOverflow; see for example 1, 2, 3, all from 2016.

If std::underlying_type<T> is well-defined but has no member type when T is not an enumeration type, then the above function template can be simplified to

template<class T>
std::enable_if_t<std::is_same_v<std::underlying_type_t<T>, int>> foo(T t);

which is shorter, more intuitive, and easier to understand.

It's worth noting that libc++ has a __sfinae_underlying_type for internal use with an implementation very similar to that described below (but with an additional helper member typedef).

Impact on the standard

This is a pure extension, as currently instantiating std::underlying_type over a non-enumeration type results in undefined behavior, typically a compile-time error.

Design alternatives

A possible alternative would be to make the nested typedef type return the type unchanged if T is not an enumeration type, matching the behavior of some other TransformationTraits such as add_lvalue_reference and remove_extent. This approach also avoids hard errors. This author, however, does not see reasons to make underlying_type_t<int> well-formed.

Another design alternative would be to support char16_t, char32_t, and wchar_t, each of which also has a "underlying type" (see [basic.fundamental]/5). And though the standard doesn't use the term, one might say that plain char similarly has an "underlying type" of either signed char or unsigned char depending on the implementation. The current specification in the standard permits implementations to support those types as an extension, though the author is not aware of any implementation that does so. In the author's opinion, the two categories are sufficiently distinct that lumping them into the same type trait would not be advisable.

The prohibition against incomplete enumeration types is left undisturbed, given the potential for ODR violations, and since the benefit from supporting such types is minimal at best.

Proposed wording

Edit the table in [meta.trans.other] as indicated:

Template Condition Comments
... ... ...
template <class T>
struct underlying_type;
T shall be a complete enumeration type (7.2) If T is an enumeration type, T shall be a complete type. If T is an enumeration type, the The member typedef type shall name the underlying type of T; otherwise, there shall be no member type.
... ... ...

Possible implementation

Given an intrinsic __underlying_type(T) (which is already needed to implement the current version of underlying_type), the implementation is trivial:

 template<class T, bool = std::is_enum_v<T>> struct _Underlying_type {};
 template<class T> struct _Underlying_type<T, true> { using type = __underlying_type(T); };

 template<class T> struct underlying_type : _Underlying_type<T> { };