Question

I overload an operator twice with the same parameter list. but with different return type:

T& operator()(par_list){blablabla}    
const T& operator()(par_list){blablabla}

So when I’m calling the () operator, which function would be called based on what preference or situation? I know that if I call () under const function it has to be the const T& one.

I’m just curious how C++ deal with such situation and how the default preference works.

Thanks

Answer 1

These functions don’t overload each other; they have the same signatures, and so the attempt to redefine the same function, which is an error. The return type is not part of a function’s signature. To overload a function, you must declare a second function with the same name, but different parameters or const/volatile qualifiers – that is, qualifiers on the function, not the return type.

(They don’t override each other either; overriding is what derived classes do to their base classes’ virtual functions).

It’s common to define a const and a non-const overload of a member function; the const overload must declare the function const, not just the return type:

T& operator()(par_list){blablabla}
const T& operator()(par_list) const {blablabla}
                              ^^^^^

Now the first will be called if you apply () to a non-const object, and the second on a const object. For example:

Thingy nc;
Thingy const c;

nc(); // calls the first (non-const) overload
c();  // calls the second (const) overload