c++ - Check if type is hashable

Question

I would like to make a type trait for checking if a particular type is hashable using the default instantiations of the standard library's unordered containers, thus if it has a valid specialization for std::hash. I think this would be a very useful feature (e.g. for using std::set as failsafe for std::unordered_set in generic code). So I, thinking std::hash is not defined for each type, started making the following SFINAE solution:

template<typename T> std::true_type hashable_helper(
    const T&, const typename std::hash<T>::argument_type* = nullptr);

template<typename T> std::false_type hashable_helper(...);

//It won't let me derive from decltype directly, why?
template<typename T> struct is_hashable 
    : std::is_same<decltype(hashable_helper<T>(std::declval<T>())),
                   std::true_type> {};

(Forgive my modest SFINAE-abilities if this is not the best solution or even wrong.)

But then I learned, that both gcc 4.7 and VC++ 2012 define std::hash for any type T, just static_asserting in the non-specialized version. But instead of compiling conditionally they (and also clang 3.1 using gcc 4.7's libstdc++) fail the assertion resulting in a compile error. This seems reasonable since I think static_asserts are not handled by SFINAE (right?), so an SFINAE solution seems not possibly at all. It's even worse for gcc 4.6 which doesn't even have a static_assert in the general std::hash template but just doesn't define its () operator, resulting in a linker error when trying to use it (which is always worse than a compile error and I cannot imagine any way to transform a linker error into a compiler error).

So is there any standard-conformant and portable way to define such a type trait returning if a type has a valid std::hash specialization, or maybe at least for the libraries static_asserting in the general template (somehow transforming the static_assert error into a SFINAE non-error)?

Answer 1

Since C++17 it is now possible to do this in a more elegant way. From cppreference about std::hash:

Each specialization of this template is either enabled ("untainted") or disabled ("poisoned"). For every type Key for which neither the library nor the user provides an enabled specialization std::hash, that specialization exists and is disabled. Disabled specializations do not satisfy Hash, do not satisfy FunctionObject, and std::is_default_constructible_v, std::is_copy_constructible_v, std::is_move_constructible_v, std::is_copy_assignable_v, std::is_move_assignable_v are all false. In other words, they exist, but cannot be used.

This meant that the STL had to remove the static_assert in C++17. Here is a working solution with 'Clang-6.0.0 -std=c++17':

#include <functional>
#include <ios>
#include <iostream>
#include <type_traits>

template <typename T, typename = std::void_t<>>
struct is_std_hashable : std::false_type { };

template <typename T>
struct is_std_hashable<T, std::void_t<decltype(std::declval<std::hash<T>>()(std::declval<T>()))>> : std::true_type { };

template <typename T>
constexpr bool is_std_hashable_v = is_std_hashable<T>::value; 

struct NotHashable {};

int main()
{
    std::cout << std::boolalpha;
    std::cout << is_std_hashable_v<int> << std::endl;
    std::cout << is_std_hashable_v<NotHashable> << std::endl;
    return 0;
}

This might for example come in handy when you use boost::hash_combine or boost::hash_range. If you include a header containing the following code sample you do not need to define boost hashes for specific types anymore.

#include <boost/functional/hash_fwd.hpp>

template <typename T, typename = std::void_t<>>
struct is_boost_hashable : std::false_type { };

template <typename T>
struct is_boost_hashable<T, std::void_t<decltype(boost::hash_value(std::declval<T>()))>> : std::true_type { };

template <typename T>
constexpr bool is_boost_hashable_v = is_boost_hashable<T>::value;  

namespace boost
{
    template <typename T>
    auto hash_value(const T &arg) -> std::enable_if_t<is_std_hashable_v<T> &&
                                                      !is_boost_hashable_v<T>, std::size_t>
    {
        return std::hash<T>{}(arg);
    }
}

Notice the is_boost_hashable_v, this is necessary to avoid ambiguity as boost already provides hashes for a lot of hashes.