使用元组作为键构建无序映射

Question

在使用Boost的C++程序中,我正在尝试构建一个无序映射,其键是双精度元组:

typedef boost::tuples::tuple<double, double, double, double> Edge;
typedef boost::unordered_map< Edge, int > EdgeMap;

但是,当我尝试使用键和值填充它时,初始化地图就完成了

EdgeMap map;
Edge key (0.0, 0.1, 1.1, 1.1);
map[key] = 1;

我遇到以下错误消息:

/usr/include/boost/functional/hash/extensions.hpp:176: error: no matching function for call to ‘hash_value(const boost::tuples::tuple<double, double, double, double, boost::tuples::null_type, boost::tuples::null_type, boost::tuples::null_type, boost::tuples::null_type, boost::tuples::null_type, boost::tuples::null_type>&)’

我认为这是因为我需要为元组键指定一个哈希函数.我怎样才能做到这一点？

编辑:

根据以下建议,我写了以下实现:

#include <boost/tuple/tuple.hpp>
#include <boost/unordered_map.hpp>

typedef boost::tuples::tuple<double, double, double, double> Edge;

struct ihash
    : std::unary_function<Edge, std::size_t>
{
    std::size_t operator()(Edge const& e) const
    {
        std::size_t seed = 0;
        boost::hash_combine( seed, e.get<0>() );
        boost::hash_combine( seed, e.get<1>() );
        boost::hash_combine( seed, e.get<2>() );
        boost::hash_combine( seed, e.get<3>() );
        return seed;
    }
};

struct iequal_to
    : std::binary_function<Edge, Edge, bool>
{
    bool operator()(Edge const& x, Edge const& y) const
    {
        return ( x.get<0>()==y.get<0>() &&
                 x.get<1>()==y.get<1>() &&
                 x.get<2>()==y.get<2>() &&
                 x.get<3>()==y.get<3>());
    }
};

typedef boost::unordered_map< Edge, int, ihash, iequal_to > EdgeMap;

int main() {

    EdgeMap map;
    Edge key (0.0, 0.1, 1.1, 1.1);
    map[key] = 1;

    return 0;
}

有可能缩短它吗？

Answer 1

实际上,你可以完美地定义一般的哈希函数boost::tuple.唯一的要求是它位于同一名称空间内,以便ADL选择它.

我真的很惊讶他们还没写过.

namespace boost { namespace tuples {

  namespace detail {

    template <class Tuple, size_t Index = length<Tuple>::value - 1>
    struct HashValueImpl
    {
      static void apply(size_t& seed, Tuple const& tuple)
      {
        HashValueImpl<Tuple, Index-1>::apply(seed, tuple);
        boost::hash_combine(seed, tuple.get<Index>());
      }
    };

    template <class Tuple>
    struct HashValueImpl<Tuple,0>
    {
      static void apply(size_t& seed, Tuple const& tuple)
      {
        boost::hash_combine(seed, tuple.get<0>());
      }
    };
  } // namespace detail

  template <class Tuple>
  size_t hash_value(Tuple const& tuple)
  {
    size_t seed = 0;
    detail::HashValueImpl<Tuple>::apply(seed, tuple);
    return seed;
  }

} }

注意:我只证明它是正确的,我还没有测试过.

Answer 2

You need a bit of front matter. Because of the underlying implementation of boost::tuples::tuple, make Edge a structure to allow the overloads to resolve correctly. Otherwise, you'll get no matches for

• boost::hash_value(const Edge &)
• operator==(const Edge &, const Edge &)

Code below:

struct Edge {
  Edge(double x1, double x2, double x3, double x4)
    : tuple(x1,x2,x3,x4) {}
  boost::tuples::tuple<double, double, double, double> tuple;
};

// XXX: less than ideal implementation!
bool operator==(const Edge &a, const Edge &b)
{
  return a.tuple.get<0>() == b.tuple.get<0>() &&
         a.tuple.get<1>() == b.tuple.get<1>() &&
         a.tuple.get<2>() == b.tuple.get<2>() &&
         a.tuple.get<3>() == b.tuple.get<3>();
}

// XXX: me too!
std::size_t hash_value(const Edge &e)
{
  std::size_t seed = 0;
  boost::hash_combine(seed, e.tuple.get<0>());
  boost::hash_combine(seed, e.tuple.get<1>());
  boost::hash_combine(seed, e.tuple.get<2>());
  boost::hash_combine(seed, e.tuple.get<3>());
  return seed;
}

typedef boost::unordered_map< Edge, int > EdgeMap;