Question

say i have the following C++ class.

class C
{
   int foo() const;
   int & foo();
};

and i just call myC.foo(), i can see using the debugger that it calls the one with the reference.

why?

thanks!

Answer 1

It’s likely because myC is a non-const value and hence the compiler is prefering the non-const method. The const method will only be preferred when accessed from a const value. For example

C value1;
value1.foo();  // int& foo();
const C value2;
value2.foo();  // int foo() const;

EDIT

Also, as Oli pointed out, overload resolution in C++ is not affected by the return type of the method. So it won’t pick one of these signatures over the other based on the way in which the value is used.