Is unordered_set<reference_wrapper<Ty>> valid?

Question

Is this valid C++ (considering the latest standard)? I'm getting compilation errors with near-top-of-tree clang/libc++ on Ubuntu 12.04. If it should be valid, I'll mail the clang-dev list with error messages and such.

#include <functional>
#include <unordered_set>

struct X
{
    int i; 
};

void f ()
{
    std::unordered_set<std::reference_wrapper<X>> setOfReferencesToX;

    // Do stuff with setOfReferencesToX
}

** As an aside, I'm tired of qualifying that the question/answer is specific to the latest standard. Could the C++ community as a whole, please start qualifying things that are specific to the old standard instead? The newer standard has been out for about a year now.

Answer 1

The problem is not specific to std::reference_wrapper<T>, but rather to the type X itself.

The issue is that std::unordered_set requires that you define hashing and equality functors for std::reference_wrapper<X>. You can pass the hash functor as second template parameter.

For example, this would work:

#include <functional> // for std::hash<int>

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

and then

std::unordered_set<std::reference_wrapper<X>, HashX> setOfReferencesToX;

Another option is to make a specialization for std::hash<X>:

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

This allows you to avoid explicitly specifying the 2nd template argument:

std::unordered_set<std::reference_wrapper<X>> setOfReferencesToX;

Concerning the equality comparison, you can fix this by providing an equality operator for class X:

struct X
{
  bool operator==(const X& rhs) const { return i == rhs.i; }
  int i; 
};

Otherwise, you can define your own functor and pass it as third template argument.