Why this code provides specialization for ALL enums for std::hash template?

Question

I'm not a pro in C++ but somehow I provided a solution while porting my MSVS 2015 C++ code to MinGW 4.9.2 to specialize std::hash class to support all enums. It there's any C++ compiler developers or C++ pro programmers here, can you explain why this specialization works, although it is Undefined Behavior according the C++ standard they say?

Link for full code with samples on Gist

#include <unordered_set>
#include <functional>

#if (defined __GNUC__) && (__GNUC__ < 6)
// Adds support of std::hash<enum T> to libstdc++.
// GCC 6 provides it out of the box.
namespace std {
    template<typename T>
    struct hash {
        constexpr size_t operator()(typename std::enable_if<std::is_enum<T>::value, T>::type s) const noexcept {
            return static_cast<size_t>(s);
        }
    };
}
#endif

Providing support for std::hash<enum T> means that all classes like std::unordered_XXX will support any enum as a key.

GCC 6.1.0 with this std::hash definition fails with error

error: redefinition of 'struct std::hash<_Tp>'

GCC 5.3.0 without this std::hash definition fails for std::unordered_set with following error:

In file included from /usr/local/gcc-5.3.0/include/c++/5.3.0/bits/hashtable.h:35:0,
                 from /usr/local/gcc-5.3.0/include/c++/5.3.0/unordered_set:47,
                 from prog.cc:1:
/usr/local/gcc-5.3.0/include/c++/5.3.0/bits/hashtable_policy.h: In instantiation of 'struct std::__detail::__is_noexcept_hash<Foo, std::hash<Foo> >':
/usr/local/gcc-5.3.0/include/c++/5.3.0/type_traits:137:12:   required from 'struct std::__and_<std::__is_fast_hash<std::hash<Foo> >, std::__detail::__is_noexcept_hash<Foo, std::hash<Foo> > >'
/usr/local/gcc-5.3.0/include/c++/5.3.0/type_traits:148:38:   required from 'struct std::__not_<std::__and_<std::__is_fast_hash<std::hash<Foo> >, std::__detail::__is_noexcept_hash<Foo, std::hash<Foo> > > >'
/usr/local/gcc-5.3.0/include/c++/5.3.0/bits/unordered_set.h:95:63:   required from 'class std::unordered_set<Foo>'
prog.cc:25:25:   required from here
/usr/local/gcc-5.3.0/include/c++/5.3.0/bits/hashtable_policy.h:85:34: error: no match for call to '(const std::hash<Foo>) (const Foo&)'
  noexcept(declval<const _Hash&>()(declval<const _Key&>()))>
                                  ^
In file included from /usr/local/gcc-5.3.0/include/c++/5.3.0/bits/move.h:57:0,
                 from /usr/local/gcc-5.3.0/include/c++/5.3.0/bits/stl_pair.h:59,
                 from /usr/local/gcc-5.3.0/include/c++/5.3.0/utility:70,
                 from /usr/local/gcc-5.3.0/include/c++/5.3.0/unordered_set:38,
                 from prog.cc:1:
/usr/local/gcc-5.3.0/include/c++/5.3.0/type_traits: In instantiation of 'struct std::__not_<std::__and_<std::__is_fast_hash<std::hash<Foo> >, std::__detail::__is_noexcept_hash<Foo, std::hash<Foo> > > >':
/usr/local/gcc-5.3.0/include/c++/5.3.0/bits/unordered_set.h:95:63:   required from 'class std::unordered_set<Foo>'
prog.cc:25:25:   required from here
/usr/local/gcc-5.3.0/include/c++/5.3.0/type_traits:148:38: error: 'value' is not a member of 'std::__and_<std::__is_fast_hash<std::hash<Foo> >, std::__detail::__is_noexcept_hash<Foo, std::hash<Foo> > >'
     : public integral_constant<bool, !_Pp::value>
...

Yakk - Adam Nevraumont · Accepted Answer · 2016-07-23T12:41:48.640

This is an ill formed program with no diagnostic required because are not allowed to provide base specializations for template types in std.

It worked because in some implementations, providing such a base specialization happens to redirect all std::hash<T> without an explicit specialization to your hash implementation. Then your operator() proceeds to only work with enums, but that is not why it works, nor does it prevent it from making your program ill formed with no diagnostic required.

Practically, it probably broke because someone SFINAE enabled std::hash with an empty base implementation in gcc 6.1: this may be mandated by some C++ standard, unsure, but it is a quality of implementation improvement if not.

The right way to do this is to create

struct enum_hash {
  template<typename T>
  constexpr
  typename std::enable_if<std::is_enum<T>::value,std::size_t>::type
  operator()(T s) const noexcept {
    return static_cast<std::size_t>(s);
  }
};

Which is a type that can hash any enum.

Now pass ,enum_hash to unordered_set<some_enum, enum_hash> like that.

Looks like there is only template struct hash; in GCC < 6.1.0. I mean there's no body for this template and this means that I providing the body by my code, am I correct? — Vyacheslav Napadovsky, Jul 23 '16 at 08:22
@slav yes. It is still an ill formed program with no diagnostic required, but the compiler generates a program with no error message, and that program (whose behaviour is no longer governed by the standard) happens to behave in a way matching what you want, due to implementation quirks. Basically, doing this is not worth it, when the cost it saves is to add a `,enum_hash` to unordered map/sets. — Yakk - Adam Nevraumont, Jul 23 '16 at 12:37

Why this code provides specialization for **ALL** enums for std::hash template?

1 Answers1

Why this code provides specialization for ALL enums for std::hash template?