Const method - repeat in implementation?

Question

I've been searching a lot for an answer and can´t find it anywere. Say i have:

class foobar{
  public:
     char foo() const;
};

, in my foobar.h

When I want to implement this class in foobar.cpp should I repeat const?:

char foobar::foo() const{
//...my code
}

Or can i do (whitout the const)

char foobar::foo() {
//...my code
}

---

If this is a duplicate I'm sorry, but no other question truly answered this.

Answer 1

Yes, you have to include the const qualifier in the definition. If you write:

class Foo
{
public:
    int f () const;
};

And in implementation file if you write:

int Foo::f () { /*...*/ }

then the compiler will emit an error saying that there is not function with signature int f () in the class. If you put the const keyword in your implementation file too, it will work.

It it possible to overload functions according to the object constness. Example:

class Foo
{
public:
    int foo ()       { std::cout << "non-const foo!" << std::endl; }
    int foo () const { std::cout << "const foo!" << std::endl; }
};

int main ()
{
    Foo f;
    const Foo cf;
    f.foo ();
    cf.foo ();
}

The output will be (as expected):

non-const foo!

const foo!

As we did with const, you can overload a function based on the object volatileness too.

Answer 2

You absolutely must include the const qualifier in the implementation.

It is possible to overload functions according to their constness. In fact this is a very important part of the language.