C++ STL <map> 允许重复对吗？
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STL <map> allows duplicate pairs?
提问by dead programmer
I have written the following code and was surprised at the output. I heard that <map>avoids collision of keys, but here it appears to allow insertion of duplicate pairs.
我编写了以下代码并对输出感到惊讶。我听说这样<map>可以避免键冲突，但在这里它似乎允许插入重复的对。
#include<iostream>
#include<map>
using namespace std;
int main()
{
map<string,char> namemap;
namemap["yogi"]='c';
namemap.insert(pair<string,char>("yogendra",'a'));
namemap.insert(pair<string,char>("yogendra",'b'));
cout<<namemap["yogendra"]<<endl;
return 0;
}
This code outputs a. You can run it on C++ Shell.
此代码输出a. 您可以在C++ Shell上运行它。
Does avoiding collision means that we cannot enter multiple pairs with same key?
避免碰撞是否意味着我们不能使用相同的密钥输入多个对？
回答by NPE
The second insertwith the same key is a no-op. It simply returns an iterator pointing to the existing element.
insert具有相同密钥的第二个是空操作。它只是返回一个指向现有元素的迭代器。
std::map::insert()has a return value, which you should check.
std::map::insert()有一个返回值，您应该检查它。
It is of type std::pair<iterator,bool>. The second element of the pair tells you whether the element has been inserted, or whether there was already an existing entry with the same key.
它的类型为std::pair<iterator,bool>。该对的第二个元素告诉您该元素是否已被插入，或者是否已经存在具有相同键的条目。
cout << namemap.insert(pair<string,char>("yogendra",'a')).second << endl;
cout << namemap.insert(pair<string,char>("yogendra",'b')).second << endl;
回答by Jay D
STL map does not allow same Keys to be used. You may want to go for multi-map for that.
STL 映射不允许使用相同的密钥。为此，您可能想要使用多地图。
回答by ashutosh
a map will not throw any compile/run time error while inserting value using duplicate key. but while inserting, using the duplicate key it will not insert a new value, it will return the same exiting value only. it will not overwrite. but in the below case it will be overwritten.
使用重复键插入值时，映射不会引发任何编译/运行时错误。但是在插入时，使用重复键不会插入新值，它只会返回相同的现有值。它不会覆盖。但在以下情况下，它将被覆盖。
map<char,int> m1;
m1.insert(pair <char, int> ('a', 40));
m1['a']=50;
cout << "a => " << m1.find('a')->second << '\n';
The result will be 50.
结果将是 50。
below example, it will not overwrite.
下面的例子，它不会覆盖。
map<char,int> m1;
m1.insert(pair <char, int> ('a', 40));
m1.insert(pair <char, int> ('a', 50));
cout << "a => " << m1.find('a')->second << '\n';
Result will be 40.
结果将是 40。
Remember map size 1 here for both the cases.
对于这两种情况，请记住这里的地图大小 1。
cout< "size = " << m1.size() << '\n';
it will be 1 in both cases.
在这两种情况下它都是 1。
