How does overloading of const and non-const functions work?

Question

The STL is full of definitions like this:

iterator begin ();
const_iterator begin () const;

As return value does not participate in overloading resolution, the only difference here is the function being const. Is this part of the overloading mechanism? What is the compiler's algorithm for resolving a line like:

vector<int>::const_iterator it = myvector.begin();

Answer 1

The compiler's "algorithm" is like this: Every member function of class X has an implicit argument of type X& (I know, most think it's X*, but the standard states, that for purposes of Overload Resolution we assume it to be a reference). For const functions, the type of the argument is const X&. Thus the algorithm, if a member function is called the two versions, const and non-const, are both viable candidates, and the best match is selected just as in other cases of overload resolution. No magic :)

Answer 2

In the example you gave:

vector<int>::const_iterator it = myvector.begin();

if myvector isn't const the non-const version of begin() will be called and you will be relying on an implicit conversion from iterator to const_iterator.