建立一个无序地图元组作为键 [英] Building an unordered map with tuples as keys

问题描述

在使用升压C ++程序，我想建立一个无序映射，其关键字是双打的元组：

In a C++ program with Boost, I am trying to build an unordered map whose keys are tuples of doubles:

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

不过初始化地图完成后确定，当我尝试用键和值来填充它。

Initializing the map completes OK, however, when I try to populate it with keys and values

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

我遇到以下错误消息：

I encounter the following error message:

/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presume这是因为我需要指定的哈希函数的元组键。我该怎么做？

I presume this is because I need to specify a hash function for the tuple keys. How can I do that?

编辑：

按照下面的建议，我写了下面的实现：

Following the suggestions below, I wrote the following implementation:

#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;
}

是否有可能缩短？

Is it possible to shorten it?

推荐答案

您需要一点点的前事。因为的boost ::元组::元组的底层实现，让边缘允许过载来解决结构正确。否则，你将得不到任何比赛为

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

• 的boost ::哈希值（常量边及放大器;）


• 运算符==（const的边缘和放大器;，常量边及放大器;）

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

code如下：

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;

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