在unordered_map中使用元组

问题描述 投票:13回答:5

我想在我的int中使用由charcharunordered_map组成的元组。我这样做:

#include <string>
#include <unordered_map>
#include <cstring>
#include <iostream>
#include <tuple>

using namespace std;

tuple <int,char,char> kk;
unordered_map<kk,int> map;

int main()
{
    map[1,"c","b"]=23;
    return 0;
}

但这给了我以下错误:

map.cpp:9:21: error: type/value mismatch at argument 1 in template parameter list     for ‘template<class _Key, class _Tp, class _Hash, class _Pred, class _Alloc> class    std::unordered_map’
map.cpp:9:21: error:   expected a type, got ‘kk’
map.cpp:9:21: error: template argument 3 is invalid
map.cpp:9:21: error: template argument 4 is invalid
map.cpp:9:21: error: template argument 5 is invalid
map.cpp:9:26: error: invalid type in declaration before ‘;’ token
map.cpp: In function ‘int main()’:
map.cpp:14:16: error: assignment of read-only location ‘"b"[map]’

我在这做错了什么?

c++ hashmap unordered-map stdtuple
5个回答
15
投票

unordered_map的模板参数如下所示:

template<

    class Key,
    class T,
    class Hash = std::hash<Key>,
    class KeyEqual = std::equal_to<Key>,
    class Allocator = std::allocator< std::pair<const Key, T> >
> class unordered_map;

std::hashnot specialized for tuples(向下滚动到库类型的标准专业化)。因此,您需要提供自己的,如下所示:

typedef std::tuple<int, char, char> key_t;

struct key_hash : public std::unary_function<key_t, std::size_t>
{
 std::size_t operator()(const key_t& k) const
 {
   return std::get<0>(k) ^ std::get<1>(k) ^ std::get<2>(k);
 }
};
// ..snip..
typedef std::unordered_map<const key_t,data,key_hash,key_equal> map_t;
//                                             ^ this is our custom hash

最后,正如本杰明林德利回答已经解决的那样,你需要使用std::make_tuple

// d is data
m[std::make_tuple(1, 'a', 'b')] = d;
auto itr = m.find(std::make_tuple(1, 'a', 'b'));

代码是从Using a std::tuple as key for std::unordered_map抓住的,这里是Live Example


12
投票

第一个错误:

map.cpp:9:21: error:   expected a type, got ‘kk’

正如错误清楚地说明的那样,模板参数需要是一种类型。 kk不是一种类型,它是一个对象。也许你想让它成为一个typedef?

typedef tuple <int,char,char> kk;
unordered_map<kk,int> map;

第二个错误:

map[1,"c","b"]=23;

这里有两个问题。首先,在值之间加上逗号不会产生元组。你需要明确它,要么调用元组类型的构造函数,要么使用返回元组的函数(例如std::make_tuple)。其次,你的元组期待字符('c','b'),而不是字符串("c","b")。

map[std::make_tuple(1,'c','b')] = 23;

7
投票

正如所指出的,std :: hash不是专门用于元组的。但是,如果你的元组由像string和int这样的标准可扩展类型组成,那么generic-hash-for-tuples-in-unordered-map-unordered-set的以下代码将自动在c ++ 11中添加这样的支持。

只需将代码粘贴到头文件中,并在需要时包含它:

#include <tuple>
// function has to live in the std namespace 
// so that it is picked up by argument-dependent name lookup (ADL).
namespace std{
    namespace
    {

        // Code from boost
        // Reciprocal of the golden ratio helps spread entropy
        //     and handles duplicates.
        // See Mike Seymour in magic-numbers-in-boosthash-combine:
        //     https://stackoverflow.com/questions/4948780

        template <class T>
        inline void hash_combine(std::size_t& seed, T const& v)
        {
            seed ^= hash<T>()(v) + 0x9e3779b9 + (seed<<6) + (seed>>2);
        }

        // Recursive template code derived from Matthieu M.
        template <class Tuple, size_t Index = std::tuple_size<Tuple>::value - 1>
        struct HashValueImpl
        {
          static void apply(size_t& seed, Tuple const& tuple)
          {
            HashValueImpl<Tuple, Index-1>::apply(seed, tuple);
            hash_combine(seed, get<Index>(tuple));
          }
        };

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

    template <typename ... TT>
    struct hash<std::tuple<TT...>> 
    {
        size_t
        operator()(std::tuple<TT...> const& tt) const
        {                                              
            size_t seed = 0;                             
            HashValueImpl<std::tuple<TT...> >::apply(seed, tt);    
            return seed;                                 
        }                                              

    };
}

2
投票

我需要地图而不是无序地图: 关键是3元组和 价值是一个4元组

看到所有的答案,我即将变成对

但是,下面对我有用:

// declare a map called map1
map <
  tuple<short, short, short>,
  tuple<short, short, short, short>
> map1;

// insert an element into map1
map1[make_tuple(1, 1, 1)] = make_tuple(0, 0, 1, 1);

// this also worked
map1[{1, 1, 1}] = { 0, 0, 1, 1 };

我正在使用visual studio community 2015 ide


0
投票

这是一种使用元组作为unordered_map的键而不使用散列特化的方法:

#include <string>
#include <tuple>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <vector>
#include <unordered_map>
using namespace std;

string fToStr(unordered_map<double,int>& dToI,float x)
{
   static int keyVal=0;
   stringstream ss;
   auto iter = dToI.find(x);
   if(iter == dToI.end()) {
      dToI[x]=++keyVal;
      ss << keyVal;
   } else {
      ss <<  iter->second;
   }
   return ss.str();
}

typedef tuple<int,char,char> TICC;
const char ReservedChar=',';
string getKey(TICC& t)
{
   stringstream ss;
   ss << get<0>(t) << ReservedChar << get<1>(t) << ReservedChar << get<2>(t);
   return ss.str();
}

int main()
{
   unordered_map< string,TICC > tupleMp;
   vector<TICC> ticc={make_tuple(1, 'a', 'b'),make_tuple(1, 'b', 'c'),make_tuple(2, 'a', 'b')};
   for(auto t : ticc)
      tupleMp[getKey(t)]=t;

   for(auto t : ticc) {
      string key = getKey(t);
      auto val = tupleMp[key];
      cout << "tupleMp[" << key << "]={" << get<0>(val) << "," << get<1>(val) <<  ","<< get<2>(val) << "} ";
   }
   cout << endl;

   //for float tuple elements use a second float to int key map 
   unordered_map< double,int > dToI;
   vector<float> v{1.234,1.234001,1.234001};
   cout << "\nfloat keys: ";
   for(float f : v)
      cout <<  setprecision(7) << f << "=" << fToStr(dToI,f) << " ";
   cout << endl;
   return 0;
}

输出是:

tupleMp[1,a,b]={1,a,b} tupleMp[1,b,c]={1,b,c} tupleMp[2,a,b]={2,a,b}

float keys: 1.234=1 1.234001=2 1.234001=2
© www.soinside.com 2019 - 2024. All rights reserved.