How to specialize std::hash::operator() for user-defined type in unordered containers?

问题

To support user-defined key types in std::unordered_set<Key> and std::unordered_map<Key, Value> one has to provide operator==(Key, Key) and a hash functor:

struct X { int id; /* ... */ };
bool operator==(X a, X b) { return a.id == b.id; }

struct MyHash {
  size_t operator()(const X& x) const { return std::hash<int>()(x.id); }
};

std::unordered_set<X, MyHash> s;

It would be more convenient to write just std::unordered_set<X> with a default hash for type X, like for types coming along with the compiler and library. After consulting

C++ Standard Draft N3242 §20.8.12 [unord.hash] and §17.6.3.4 [hash.requirements],
Boost.Unordered
g++ include\\c++\\4.7.0\\bits\\functional_hash.h
VC10 include\\xfunctional
various related questions in Stack Overflow

it seems possible to specialize std::hash<X>::operator():

namespace std { // argh!
  template <>
  inline size_t 
  hash<X>::operator()(const X& x) const { return hash<int>()(x.id); } // works for MS VC10, but not for g++
  // or
  // hash<X>::operator()(X x) const { return hash<int>()(x.id); }     // works for g++ 4.7, but not for VC10 
}

Given compiler support for C++11 is yet experimental---I did not try Clang---, these are my questions:

Is it legal to add such a specialization to namespace std? I have mixed feelings about that.
Which of the std::hash<X>::operator() versions, if any, is compliant with C++11 standard?
Is there a portable way to do it?

回答1:

You are expressly allowed and encouraged to add specializations to namespace std*. The correct (and basically only) way to add a hash function is this:

namespace std {
  template <> struct hash<Foo>
  {
    size_t operator()(const Foo & x) const
    {
      /* your code here, e.g. "return hash<int>()(x.value);" */
    }
  };
}

(Other popular specializations that you might consider supporting are std::less, std::equal_to and std::swap.)

_{*) as long as one of the involved types is user-defined, I suppose.}

回答2:

My bet would be on the Hash template argument for the unordered_map/unorder_set/... classes:

#include <unordered_set>
#include <functional>

struct X 
{
    int x, y;
    std::size_t gethash() const { return (x*39)^y; }
};

typedef std::unordered_set<X, std::size_t(*)(const X&)> Xunset;
typedef std::unordered_set<X, std::function<std::size_t(const X&)> > Xunset2;

int main()
{
    auto hashX = [](const X&x) { return x.gethash(); };

    Xunset  my_set (0, hashX);
    Xunset2 my_set2(0, hashX); // if you prefer a more flexible set typedef
}

Of course

hashX could just as well be a global static function
in the second case, you could pass that
- the oldfashioned functor object (struct Xhasher { size_t operator(const X&) const; };)
- std::hash<X>()
- any bind expression satisfying the signature -

回答3:

@Kerrek SB has covered 1) and 3).

2) Even though g++ and VC10 declare std::hash<T>::operator() with different signatures, both library implementations are Standard compliant.

The Standard does not specify the members of std::hash<T>. It just says that each such specialization must satisfy the same "Hash" requirements needed for the second template argument of std::unordered_set and so on. Namely:

Hash type H is a function object, with at least one argument type Key.
H is copy constructible.
H is destructible.
If h is an expression of type H or const H, and k is an expression of a type convertible to (possibly const) Key, then h(k) is a valid expression with type size_t.
If h is an expression of type H or const H, and u is an lvalue of type Key, then h(u) is a valid expression with type size_t which does not modify u.