unordered_map构造函数错误（等于模板函数） [英] unordered_map constructor error (equal_to templated function)

问题描述

我想我可以有一个指向一个完全专门的模板函数的指针，但下面的代码不编译（MSVC2012）

I thought I could have a pointer to a fully-specialized template function, but the following code isn't compiling (MSVC2012)

#include <iostream>
#include <string>
#include <unordered_map>
#include <algorithm>

using namespace std;

unsigned long hashing_func(string key)
{
    unsigned long hash = 0;
    for(int i=0; i<key.size(); i++)
    {
        hash += (71*hash + key[i]) % 5;
    }
    return hash;
}

bool key_equal_fn2(string t1, string t2)
{
    return t1 == t2;
}


template<class T> bool key_equal_fn(T t1, T t2)
{
    return t1 == t2;
}

template <> bool key_equal_fn<string>(string t1, string t2)
{
    return !(t1.compare(t2));
}

int main ()
{
    unordered_map<string, string>::size_type n = 5;
    unordered_map<string, string> mymap(n, (const std::hash<string> &)hashing_func, (const std::equal_to<string> &)(key_equal_fn<string>)) ;

    mymap["paul"] = "jenna";
    mymap["frank"] = "ashley";


    return 0;
}

构造函数行返回以下错误：

The constructor line is returning the following error:

错误C2440：'type cast'：无法从'bool（__cdecl *）（T，T）'
转换为'const std :: equal_to& _Ty>&'

error C2440: 'type cast' : cannot convert from 'bool (__cdecl *)(T,T)' to 'const std::equal_to<_Ty> &'

推荐答案

两者 hashing_func 和 key_equal_fn 应该是函子对象（而不是函数）。此外，它们的类型必须提供给 unordered_map 模板，即地图应该具有以下类型：

Both hashing_func and key_equal_fn should be functor objects (and not functions). In addition, their types must be provided to the unordered_map template, that is, the map should have this type:

unordered_map<string, string, hashing_func, key_equal_fn>

其中 hashing_func 和 key_equal_fn 是函数类：

struct hashing_func {
    unsigned long operator()(const string& key) const {
        unsigned long hash = 0;
        for(size_t i=0; i<key.size(); i++)
            hash += (71*hash + key[i]) % 5;
        return hash;
    }
};

struct key_equal_fn {
    bool operator()(const string& t1, const string& t2) const {
        return !(t1.compare(t2));
    }
};

然后，以这种方式定义 mymap ：

Then, mymap is defined in this way:

typedef unordered_map<string, string, hashing_func, key_equal_fn> MapType;
MapType::size_type n = 5;
MapType mymap(n, hashing_func(), key_equal_fn());

或者， hashing_func 和/ c $ c> key_equal_fn 可以是函数，但是你必须将它们包装到 std :: function 对象中。也就是说，

Alternatively, hashing_func and/or key_equal_fn can be functions but you have to wrapp them into std::function objects. That is,

unsigned long hashing_func(const string& key) {
    unsigned long hash = 0;
    for(size_t i=0; i<key.size(); i++)
      hash += (71*hash + key[i]) % 5;
    return hash;
}

bool key_equal_fn(const string& t1, const string& t2){
  return !(t1.compare(t2));
}

并定义 mymap

typedef unordered_map<string, string,
    std::function<unsigned long(const string&)>,
    std::function<bool(const string&, const string&)>> MapType;

MapType::size_type n = 5;
MapType mymap(n, hashing_func, key_equal_fn);

如果你愿意，你可以使用lambdas并禁止写两个函数或函子类： p>

If you wish, you can use lambdas and refrain from writing the two funcions or functor classes:

typedef unordered_map<string, string,
    std::function<unsigned long(const string&)>,
    std::function<bool(const string&, const string&)>> MapType;

MapType mymap(n,
  [](const string& key) -> unsigned long {
      unsigned long hash = 0;
      for(size_t i=0; i<key.size(); i++)
        hash += (71*hash + key[i]) % 5;
      return hash;
  },
  [](const string& t1, const string& t2) {
       return !(t1.compare(t2));
  });

最后，我最喜欢的是 all-lambdas

Finally, my favorite is an all-lambdas solution

auto hashing_func = [](const string& key) -> unsigned long {
    unsigned long hash = 0;
    for(size_t i=0; i<key.size(); i++)
        hash += (71*hash + key[i]) % 5;
    return hash;
};

auto key_equal_fn = [](const string& t1, const string& t2) {
    return !(t1.compare(t2));
};

typedef unordered_map<string, string,
    decltype(hashing_func), decltype(key_equal_fn)> MapType;

MapType::size_type n = 5;
MapType mymap(n, hashing_func, key_equal_fn);

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