對於非庫類型，無序的key_type需要散列<>專業化？

Question

我試圖用xercesc :: XMLUri作爲關鍵類型創建一個std :: unordered_map。

#include <unordered_map> 
#include "xercesc/util/XMLUri.hpp" 

int main() 
{ 
     std::unordered_map<xercesc::XMLUri,xercesc::XMLUri> uriMap; 
} 

導致以下：

clang++ -std=c++11 -O0 -emit-llvm -g3 -Wall -c -fmessage-length=0 -I/usr/include ../xx.cpp 
In file included from ../xx.cpp:1: 
In file included from /usr/bin/../lib/gcc/i686-linux-gnu/4.7/../../../../include/c++/4.7/unordered_map:43: 
/usr/bin/../lib/gcc/i686-linux-gnu/4.7/../../../../include/c++/4.7/bits/functional_hash.h:59:7: error: static_assert failed "std::hash is not specialized for this type" 
    static_assert(sizeof(_Tp) < 0, 
^   ~~~~~~~~~~~~~~~ 
/usr/bin/../lib/gcc/i686-linux-gnu/4.7/../../../../include/c++/4.7/bits/unordered_map.h:45:32: note: in instantiation of template class 'std::hash<xercesc_3_1::XMLUri>' requested here 
         integral_constant<bool, !__is_final(_Hash)>, 
               ^
/usr/bin/../lib/gcc/i686-linux-gnu/4.7/../../../../include/c++/4.7/bits/unordered_map.h:263:14: note: in instantiation of default argument for '__unordered_map<xercesc_3_1::XMLUri, xercesc_3_1::XMLUri, std::hash<xercesc_3_1::XMLUri>, std::equal_to<xercesc_3_1::XMLUri>, std::allocator<std::pair<const xercesc_3_1::XMLUri, xercesc_3_1::XMLUri> > >' required here 
: public __unordered_map<_Key, _Tp, _Hash, _Pred, _Alloc> 
     ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 
../xx.cpp:6:54: note: in instantiation of template class 'std::unordered_map<xercesc_3_1::XMLUri, xercesc_3_1::XMLUri, std::hash<xercesc_3_1::XMLUri>, std::equal_to<xercesc_3_1::XMLUri>, std::allocator<std::pair<const xercesc_3_1::XMLUri, xercesc_3_1::XMLUri> > >' requested here 
    std::unordered_map<xercesc::XMLUri,xercesc::XMLUri> uriMap; 

我知道C++ 0x中的無序容器只供應hash<>特對一些庫類型。如何爲xercesc::XMLUri創建所需的hash<xercesc::XMLUri>專業化？

編輯：我想出了這個。這看起來合理嗎？

#include "xercesc\util\XMLUri.hpp" 
#include <string> 

namespace std 
{ 

    size_t hash<xercesc::XMLUri>::operator()(const xercesc::XMLUri& uri) const 
    { 
     return hash<std::wstring>()(uri.getUriText()); 
    } 
} 

Answer 1

差不多。它應該是這樣的（感謝@jogojapan您指出缺少的類型定義！）：

#include <string> 
#include <functional> 

namespace std 
{ 
    template <> struct hash<xercesc::XMLUri> 
    { 
     typedef size_t result_type; 
     typedef xercesc::XMLUri argument_type; 

     size_t operator()(xercesc::XMLUri const & uri) const noexcept 
     { 
      return hash<wstring>()(uri.getUriText()); 
     } 
    }; 
} 

Answer 2

std::hash是一個結構，就必須專注整個結構，而不僅僅是功能，那麼你的專業模板的方式也是錯誤的：

namespace std 
{ 
    template <> 
    struct hash<xercesc::XMLUri> 
    { 
     size_t operator()(const xercesc::XMLUri& uri) const 
     { 
      return hash<std::wstring>()(uri.getUriText()); 
     } 
    }; 
}