What does iterator->second mean?
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::pair
s. 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
.
The type of the elements of an std::map
(which is also the type of an expression obtained by dereferencing an iterator of that map) whose key is K
and value is V
is std::pair<const K, V>
- the key is const
to prevent you from interfering with the internal sorting of map values.
std::pair<>
has two members named first
and second
(see here), with quite an intuitive meaning. Thus, given an iterator i
to a certain map, the expression:
i->first
Which is equivalent to:
(*i).first
Refers to the first (const
) element of the pair
object pointed to by the iterator - i.e. it refers to a key in the map. Instead, the expression:
i->second
Which is equivalent to:
(*i).second
Refers to the second element of the pair
- i.e. to the corresponding value in the map.