Function overloading — two functions only differ by a default parameter

Question

class A{
    public:
        void foo(int x)
        {
            cout << \"foo with one\\n\";
        }

        void foo(int x, int y=10)
        {
                 


        

Answer 1

I think you can't. Because function/operator overloading is resolved by compiler at compile time. So overloading a function just by providing a default argument will cause ambiguity and compiler error.