Fun with function overloading

If you cast b to const Blah& then you can call only methods declared as const, i.e. that cannot modify the internal state of the object. In this case the choice is for const Thing* GetThing() const.

But it will continue to work if you write:

class Thing
{
public:
   Thing() {}
...
};

class Blah
{
public:
  Thing* GetThing() const;

private:

   Thing* GetThing();
};


void anotherclass::function(Blah& b)
{
   const Thing* thing = b.GetThing();
}

This will continue to work because is safe to implicitly add the const attribute to a variable, whereas it is not safe the opposite.

In your code, the compiler choose which version of your method to use depending on the fact that b is const or not.

Thanks to those who replied.

From the c++ spec...

13.3.1.4:
For non-static member functions, the type of the implicit object parameter is reference to cv X where X is
the class of which the function is a member and cv is the cv-qualification on the member function declara-
tion. [Example: for a const member function of class X, the extra parameter is assumed to have type ref-
erence to const X. ]

Then according to 13.3.3.1.1, the const variant would count as a "conversion", and thus be a worse fit.