2011-12-17 252 views
1

我想创建一个充满字符串作为键和整数值的映射。当我尝试使用它进行搜索时,问题就开始了。有人能告诉我我错了哪里?我是否在同一个陈述中有两个地图?STL映射在搜索尝试时抛出错误

invale is "ale" 
    roomno = 2; 
    // roomlist is a map 
    // rinventory is another map 

    if(roomlist[roomno].rinventory.find(invale) != map<string, int>::end()); 

我得到的错误如下。什么重载功能?这确实是一个漫长的错误。

error C2668: 'std::_Tree<_Traits>::end' : ambiguous call to overloaded function 
1>  with 
1>  [ 
1>   _Traits=std::_Tmap_traits<std::string,int,std::less<std::string>,std::allocator<std::pair<const std::string,int>>,false> 
1>  ] 
1>  c:\program files\microsoft visual studio 9.0\vc\include\xtree(569): could be 'std::_Tree<_Traits>::const_iterator std::_Tree<_Traits>::end(void) const' 
1>  with 
1>  [ 
1>   _Traits=std::_Tmap_traits<std::string,int,std::less<std::string>,std::allocator<std::pair<const std::string,int>>,false> 
1>  ] 
1>  c:\program files\microsoft visual studio 9.0\vc\include\xtree(564): or  'std::_Tree<_Traits>::iterator std::_Tree<_Traits>::end(void)' 
1>  with 
1>  [ 
1>   _Traits=std::_Tmap_traits<std::string,int,std::less<std::string>,std::allocator<std::pair<const std::string,int>>,false> 
1>  ] 
1>  while trying to match the argument list '(void)'  

在此先感谢您。

回答

3

尝试将其更改为:

if(roomlist[roomno].rinventory.find(invale) != roomlist[roomno].rinventory.end()); 
+0

谢谢您的帮助 – 2011-12-17 09:06:03

0

应该

if(roomlist[roomno].rinventory.find(invale) != roomlist[roomno].rinventory.end()); 
0

map::end()方法也不是一成不变的。

正确的方法是这样的:

map<string, int>::iterator it = roomlist[roomno].rinventory.find(invale); 
if(it != roomlist[roomno].rinventory.end()) 
    // do stuff 
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