How to secure CRTP against providing wrong superclass?

Question

template <class Derived>
class Base {
};

class Derived : public Base<Derived> {
};

What would be a good way to make the code robust another copy-paste omissions, so that the following snippet throws a compile-time error:

class AnotherDerived : public Base<Derived> {
};

I'm using Visual C++ 2013.

Give `Base` a private dtor and then make `Derived` a friend? — T.C., Feb 16 '15 at 19:36
Maybe the constructor Base(Derived&) {} - but that is close to being paranoid. — , Feb 16 '15 at 19:38
@DieterLücking: Thought about that, but this messes up the interface, too... — krlmlr, Feb 16 '15 at 19:49
@T.C.: Nice idea -- but is this possible without giving `Derived` access to other private members of `Base`? — krlmlr, Feb 16 '15 at 19:50
@krlmlr: You can add another inheritance layer to hold the members that need to remain private. Using composition should also be possible. (Remember friendship is not transitive) — Ben Voigt, Feb 16 '15 at 19:52
@BenVoigt: Thanks a lot, a second inheritance layer did the trick for me. — krlmlr, Feb 16 '15 at 20:08
The 'duplicate' answer is pre-C++11 and hence not appropriate. I vote for re-opening. Moreorver, the answers given are not fully satisfactory. — Walter, Nov 19 '15 at 14:39
@Walter: Can you suggest a better solution using C++11 features? — krlmlr, Nov 20 '15 at 11:50

score 4 · Accepted Answer · answered Feb 16 '15 at 19:46

4

Make Base's destructor private, and then make Derived a friend of Base<Derived>:

template <class Derived>
class Base {
    private: ~Base() = default;
    friend Derived;
};

class Derived : public Base<Derived> {
};

This does not actually make doing

class AnotherDerived : public Base<Derived> {
};

illegal, but any attempt to actually construct an AnotherDerived will fail.

answered Feb 16 '15 at 19:46

T.C.

Thanks, this works for me -- I now have a class `BaseImp` from which `Base` inherits in order to avoid exposing private members to the pseudo-friend. – krlmlr Feb 16 '15 at 20:10

score 1 · Answer 2 · answered Feb 16 '15 at 19:44

1

You can static_assert that the argument derives from Base<Argument>, but that's as far as you can go.

answered Feb 16 '15 at 19:44

Puppy

1

Where would you do that? Inside the base class as a member-declaration, that information might not be available (yet) -- in fact, the Standard requires the second argument of `is_base_of` is a complete type. Inside a member function, an instantiation of that member function might be required to actually perform the check. – dyp Feb 16 '15 at 19:48
@dyp: Would the check always be done if performed in the constructor of `Base`? – krlmlr Feb 16 '15 at 20:11
I'm not too familiar with static assertions -- could you please give an example? – krlmlr Feb 16 '15 at 20:11
@krlmlr As far as I know, a `static_assert` inside a constructor of `Base` is only checked if that constructor is instantiated (e.g. if it is called). Although the error message might be better, I think the requirements for the `static_assert` check to be performed are higher than the check T.C. proposed. – dyp Feb 16 '15 at 20:21

2 Answers2