Const method overloading

Question

I am wandering why C++ chooses to call non const methods on non const object for overloaded methods differing only by const method signature, Namely:

#include<iostream>

class Foo
{
 public:
  Foo() {}
  int bar() const { std::cout<<"const version called"<<std::endl; }
  int bar() { std::cout<<"version called"<<std::endl; }
};

int main()
{
  Foo f;
  f.bar();
  const Foo g;
  g.bar();
  return 0;
}

I understand that for g object, being const, the const version of bar is called. But how about f? The output is

version called
const version called

Thanks for your insights.

Answer 1

This is how overload resolution works.

First, a list of candidate functions is constructed. In your case it consists of two functions:

int Foo::bar() const;
int Foo::bar();

Each function in the list is converted to accept this argument roughly like this:

int Foo_bar(const Foo* this);
int Foo_bar(Foo* this);

Then the choice is made according to the "Best viable function" (13.3.3) part of the C++ standard, which says, in short, that the best function is the one requiring the least number of conversions for its arguments. So if you have Foo f; and call f.bar(), the best function is the one accepting non-const this, because it requires no conversions, whereas the const version would require a qualification conversion for this from Foo* to const Foo*.