C ++：从unordered_map向std :: cout输出自定义哈希键值时出错

Question

I am trying to build an std::unordered_map with a custom type as a key. The custom type is a simple std::vector<double> . The idea is that it will function as a convenient container for 2D points on a grid. Everything is working correctly except for outputting the hashed key. Here is a sample I put together to illustrate the idea:

#include <iostream>
#include <vector>
#include <unordered_map>
#include <boost/functional/hash.hpp>
#include <chrono>

namespace std
{
    template<typename Container>
    struct hash {
            std::size_t operator()(Container const& v) const
            {
                return boost::hash_range(v.begin(), v.end());
            }
    };
}

int main()
{

    std::unordered_map<std::vector<double>, double> test;

    unsigned long t = (unsigned long) std::chrono::system_clock::now().time_since_epoch().count();
    std::srand(t);
    for (uint i = 0; i < 100 ; ++i)
    {
        double d1 = i/200.0;
        double d2 = i/200.0;
        std::vector<double> v({d1, d2});
        test[v] = d1;
    }

    std::cout << "Size:" << test.size() << std::endl;
    for (const auto& it : test )
    {
        std::cout << it.first << ":" << it.second << std::endl;
    }

    return 0;
}

The hash specialisation template is courtesy of another SO thread. The trouble is that g++ spits out the following error when I try to compile the above:

cannot bind 'std::ostream {aka std::basic_ostream<char>}' lvalue to 'std::basic_ostream<char>&&'
   std::cout << it.first << ":" << it.second << std::endl;
                   ^

It is obvious that it stumbles on it.first . The code compiles and runs correctly if I remove it.first . I understand that the output will not be a vector of doubles. I did look around SO for quite a while, but I couldn't find a definitive answer on how to std::cout the hash value from an unordered map with a custom key type. Any feedback will be highly appreciated.

Thank you in advance!

Edit:

Thank you everyone for your input. This was my first encounter with non-primitive types as hashed keys, so I had the wrong idea about how the key/value pairs were stored (I assumed that the hashed value is the key, whereas in fact it is the actual custom type).

Answer 1

The value_type of an unordered_map<K,V> is pair<const K, V> . That's what you get when you iterate over it with a range-for. There's no operator<< overload for vector s, causing the error you see.

---

namespace std
{
    template<typename Container>
    struct hash {
            std::size_t operator()(Container const& v) const
            {
                return boost::hash_range(v.begin(), v.end());
            }
    };
}

This isn't a specialization of std::hash . It's a redefinition of the primary template, which in your case only compiled by pure happenstance. (The implementation will have to leave the primary std::hash template undefined, and will have to actually declare hash in the std namespace and not an inlined namespace. Your code breaks up completely on libc++ , for example.)

A specialization would look like

namespace std
{
    // full specialization
    template<>
    struct hash<Foo> {
    //         ^^^^^
            std::size_t operator()(Foo const& v) const
            {
                // ...
            }
    };

    // partial specialization
    template<typename T>
    struct hash<Bar<T>>{
    //         ^^^^^^^^
            std::size_t operator()(Bar<T> const& v) const
            {
                // ...
            }
    };
}

Note the explicit template argument list following hash . That's the indication that this is a specialization.

It is illegal to specialize std::hash for std::vector<double> anyway, because it doesn't depend on a user-defined type. Writing your own hasher is easy:

struct container_hasher {
    template<typename Container>
    std::size_t operator()(Container const& v) const
    {
        using std::begin; 
        using std::end;
        return boost::hash_range(begin(v), end(v));
    }
};

Note that I templated the operator() instead of the type itself - this makes writing the hasher type easier. The using followed by an unqualified call enables ADL for begin and end .

And then the definition of test becomes

std::unordered_map<std::vector<double>, double, container_hasher> test;

Answer 2

As far as I am aware, there is no standard interface for exposing the hash value (as opposed to the hashing function) from std::unordered_map .

As you have seen, dereferencing a std::unordered_map<Key,V>::iterator yields something convertible to std::unordered_map<Key,V>::value_type , which in turn is a std::pair<const Key,V> representing a (key,value) pair, rather than a (hashed key,value) pair.

Correspondingly, it.first is giving you a std::vector<double> rather than a std::size_t .