In C++, what is the type of a std::map<>::iterator?
We know that an object it of type std::map<A,B>::iterator has an overloaded operator -> which returns a std::pair<A,B>*, and that the std::pair<> has a first and second member.
But, what do these two members correspond to, and why do we have to access the value stored in the map as it->second?
I'm sure you know that a std::vector<X> stores a whole bunch of X objects, right? But if you have a std::map<X, Y>, what it actually stores is a whole bunch of std::pair<const X, Y>s. That's exactly what a map is - it pairs together the keys and the associated values.
When you iterate over a std::map, you're iterating over all of these std::pairs. When you dereference one of these iterators, you get a std::pair containing the key and its associated value.
std::map<std::string, int> m = /* fill it */;
auto it = m.begin();
Here, if you now do *it, you will get the the std::pair for the first element in the map.
Now the type std::pair gives you access to its elements through two members: first and second. So if you have a std::pair<X, Y> called p, p.first is an X object and p.second is a Y object.
So now you know that dereferencing a std::map iterator gives you a std::pair, you can then access its elements with first and second. For example, (*it).first will give you the key and (*it).second will give you the value. These are equivalent to it->first and it->second.
