C++ STL 映射，std::pair 作为键
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C++ STL map , std::pair as the key
提问by Avinash
This is how I have defined by map.
这就是我通过地图定义的方式。
std::map<std::pair<std::string,std::string>, int> edMap;
I am confuse on how to insert values , I am getting compilation error always. This is how I am trying to insert.
我对如何插入值感到困惑，我总是收到编译错误。这就是我试图插入的方式。
std::pair<std::string,std::string> key;
edMap.insert(key,d);
Compilation error is
编译错误是
1>------ Build started: Project: spellsuggest, Configuration: Debug Win32 ------
1>Compiling...
1>breathalyzer.cpp
1>d:\personal\spellsuggest\spellsuggest\breathalyzer.cpp(70) : error C2664: 'std::_Tree<_Traits>::iterator std::_Tree<_Traits>::insert(std::_Tree<_Traits>::iterator,const std::pair<_Ty1,_Ty2> &)' : cannot convert parameter 1 from 'std::pair<_Ty1,_Ty2>' to 'std::_Tree<_Traits>::iterator'
1> with
1> [
1> _Traits=std::_Tmap_traits<std::pair<std::string,std::string>,int,std::less<std::pair<std::string,std::string>>,std::allocator<std::pair<const std::pair<std::string,std::string>,int>>,false>,
1> _Ty1=const std::pair<std::string,std::string>,
1> _Ty2=int
1> ]
1> and
1> [
1> _Ty1=std::string,
1> _Ty2=std::string
1> ]
1> and
1> [
1> _Traits=std::_Tmap_traits<std::pair<std::string,std::string>,int,std::less<std::pair<std::string,std::string>>,std::allocator<std::pair<const std::pair<std::string,std::string>,int>>,false>
1> ]
1> No user-defined-conversion operator available that can perform this conversion, or the operator cannot be called
1>Build log was saved at "file://d:\personal\spellsuggest\spellsuggest\Debug\BuildLog.htm"
1>spellsuggest - 1 error(s), 0 warning(s)
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========
回答by Kerrek SB
Let's try:
咱们试试吧：
typedef std::pair<std::string, std::string> my_key_type;
typedef std::map<my_key_type, int> my_map_type;
my_map_type m;
m.insert(my_map_type::value_type(my_key_type("A", "B"), 43));
Observe that the map's value_typeis always std::pair<const key_type, mapped_type>, so in your case it's std::pair<my_key_type, int>-- a pair whose first member is itself a pair!
观察地图value_type总是std::pair<const key_type, mapped_type>，所以在你的情况下它是std::pair<my_key_type, int>- 一对，其第一个成员本身就是一对！
With that in mind you can alternatively use make_pair:
考虑到这一点，您可以选择使用make_pair：
m.insert(std::make_pair(my_key_type("C", "D"), -5));
Finally, as Sven points out, there may or may not be a comparison operator for pairs (I think there is, though); so if there isn't, you have to write one yourself. Lexicographic comparison on the two elements should do. Sophie awaits :-)
最后，正如 Sven 指出的那样，可能有也可能没有对的比较运算符（不过我认为有）；所以如果没有，你必须自己写一个。应该对这两个元素进行字典比较。苏菲在等着:-)
(Here's the lexicographic pair comparison; you don't need to write this, it's already there:)
（这是字典对比较；你不需要写这个，它已经存在了:)
template<typename S, typename T>
bool operator<(const std::pair<S, T> & a, const std::pair<S, T> & b)
{
return (a.first < b.first) || (a.first == b.first && a.second < b.second);
}
回答by Sven
The insert method takes a full pair of the map's type, so you have to do this:
insert 方法采用完整的一对映射类型，因此您必须执行以下操作：
edMap.insert(make_pair(key, d));
回答by Rob?
Note that this:
请注意：
std::pair<std::string,std::string> key;
edMap.insert(make_pair(key,d));
will will fail to insert anything if there is already a key with the same value present.
如果已经存在具有相同值的键，则将无法插入任何内容。
This, on the other hand:
另一方面，这：
std::pair<std::string,std::string> key;
edMap[key] = d;
will either create a new item in the map, or overwrite the previous value, if one existed.
将在地图中创建一个新项目，或者覆盖以前的值（如果存在）。
