Same function with const and without – When and why?

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But why would you have both functions in one and the same class definition?

Having both allows you to:

  • call the function on a mutable object, and modify the result if you like; and
  • call the function on a const object, and only look at the result.

With only the first, you couldn’t call it on a const object. With only the second, you couldn’t use it to modify the object it returns a reference to.

And how does the compiler distinguish between these?

It chooses the const overload when the function is called on a const object (or via a reference or pointer to const). It chooses the other overload otherwise.

I believe that the second f() (with const) can be called for non-const variables as well.

If that were the only overload, then it could. With both overloads, the non-const overload would be selected instead.

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