How does the compiler handle overloaded function call operators in functors?

Question

Say I define, instantiate, and use an adder functor like so:

class SomeAdder {
    public:
        SomeAdder(int init_x): x(init_x) {}
        void operator()(int num) { cout << x + num <<endl; }
    private:
        int x;
};

SomeAdder a = SomeAdder (3);
a(5); //Prints 8

SomeAdder b(5);
b(5); //Prints 10

The constructor and the overloaded () operator are both called using double parenthesis and have the same types of parameters. How would the compiler determine which function to use during the instantiations of SomeAdder and the "function calls", as to implement the correct behavior? The answer seems like it would be obvious on the surface, but I just can't wrap my head around this thought.

Thanks for your time!

Answer 1

Every time an instance of a class is created the constructor method is called. Compiler for sure can determine constructor by its name. So it will be called first and operator () will be second.

Answer 2

Your example compares constructor and member function which overloads operator(). The compiler knows which one to call and when. It is pretty much simple:

• When an object is to be constructed, the constructor is called.

• The member function is invoked on an already constructed object. In your case, the member function is operator().

That means, they're invoked in entirely different contexts. There is no ambiguity, no confusion.