[nemetroid]

Close, it is a std::pair, but it differs in constness. Iterating a std::map<K, V> yields std::pair<const K, V>, so you have:

  std::pair<const std::string, int>

vs

  std::pair<std::string, int>

[1718627440]

And what does casting const change, that would involve runtime inefficiencies?

[gpderetta]

It is not a cast. std::pair<const std::string, ...> and std::pair<std::string,...> are different types, although there is an implicit conversion. So a temporary is implicitly created and bound to the const reference. So not only there is a copy, you have a reference to an object that is destroyed at end of scope when you might expect it to live further.

[nemetroid]

Each entry in the map will be copied. In C++, const T& is allowed to bind to a temporary object (whose lifetime will be extended). So a new pair is implicitly constructed, and the reference binds to this object.