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std::hash specialization using sfinae?

As an exercise I was trying to see if I could use SFINAE to create a std::hash specialization for std::pair and std::tuple when all of its template parameters are of an unsigned type. I have a little experience with them, but from what I understand the hash function needs to have already been templated with a typename Enabled = void for me to add a specialization. I'm not really sure where to go from here. Here's an attempt which doesn't work.

#include <functional>
#include <type_traits>
#include <unordered_set>
#include <utility>

namespace std {
template <typename T, typename Enabled = void>
struct hash<std::pair<T, T>, std::enable_if_t<std::is_unsigned<T>::value>>
{
    size_t operator()(const std::pair<T, T>& x) const
    {
        return x;
    }
};
}; // namespace std


int
main(int argc, char ** argv)
{
    std::unordered_set<std::pair<unsigned, unsigned>> test{};
    return 0;
}

Error:

hash_sfinae.cpp:7:42: error: default template argument in a class template partial specialization
template <typename T, typename Enabled = void>
                              ^
hash_sfinae.cpp:8:8: error: too many template arguments for class template 'hash'
struct hash<std::pair<T, T>, std::enable_if_t<std::is_unsigned<T>::value>>

It's about what I expected because I'm trying to extend the template parameters to hash... But I'm not sure the technique to handle these cases then. Can someone help me understand?

like image 604
Brian Rodriguez Avatar asked Jul 03 '15 19:07

Brian Rodriguez


1 Answers

You are not supposed to specialize std::hash for types that doesn't depend on a type you defined yourself.

That said, this hack might work:

template<class T, class E>
using first = T;

template <typename T>
struct hash<first<std::pair<T, T>, std::enable_if_t<std::is_unsigned<T>::value>>>
{
    size_t operator()(const std::pair<T, T>& x) const
    {
        return x;
    }
};

Really, though, don't do this. Write your own hasher.

like image 111
T.C. Avatar answered Nov 16 '22 10:11

T.C.