问题
As known the following code is used to iterate map in C++
for (std::map<char,int>::iterator it=mymap.begin(); it!=mymap.end(); ++it)
{
std::cout << itr->first << " => " << itr->second << '\n';
}
Where itr is declared as std::map::iterator. The members first and second are declared neither in std::map nor in std::iterator. Then how is it available for access?
回答1:
The elements of an std::map are std::pair<key_type, mapped_type>, so de-referencing a map iterator gives you a reference to one of these.
It is the std::pair class template which has first and second members.
回答2:
The basic idea behind iterators is that they are "magical" objects used to access data, that behave like pointers do on an array - i.e. you use arithmetic operators (e.g. ++ and --) to move around and you dereference (using * and ->) to access the data.
So, itr is "like" a pointer to an std::pair<char, int>, so you can access the data dereferencing it via the * operator (which yields the key/value pair) or with the -> operator, as in your example.
来源：https://stackoverflow.com/questions/22061866/c-iterating-map