Question

Is there any ways to define a type with a same name for classes in inheritance relationship by using CTRP? I tried the following code but got error: member 'ptr_t' found in multiple base classes of different types from clang++.

#include <iostream>
#include <tr1/memory>

template <typename T> class Pointable {
public:
    // define a type `ptr_t` in the class `T` publicly
    typedef std::tr1::shared_ptr<T> ptr_t;
};

class Parent : public Pointable<Parent> {
public:
    Parent() {
        std::cout << "Parent created" << std::endl;
    }

    ~Parent() {
        std::cout << "Parent deleted" << std::endl;
    }
};

class Child : public Parent,
              public Pointable<Child> {
public:
    Child() {
        std::cout << "Child created" << std::endl;
    }

    ~Child() {
        std::cout << "Child deleted" << std::endl;
    }
};

int main(int argc, char** argv)
{
    Child::ptr_t child_ptr(new Child());
    Parent::ptr_t parent_ptr(new Parent());

    return 0;
}

Of course, the following one is OK (but it’s redundant and go against the DRY principle).

class Parent {
public:
    typedef std::tr1::shared_ptr<Parent> ptr_t;

    Parent() {
        std::cout << "Parent created" << std::endl;
    }

    ~Parent() {
        std::cout << "Parent deleted" << std::endl;
    }
};

class Child : public Parent {
public:
    typedef std::tr1::shared_ptr<Child> ptr_t;

    Child() {
        std::cout << "Child created" << std::endl;
    }

    ~Child() {
        std::cout << "Child deleted" << std::endl;
    }
};

If there is no ways to achieve this behavior by using CRTP, why that is prohibited?

Answer 1

Your problem has nothing to do with the CRTP, but with multiple inheritance. Child inherits ptr_t from both its base classes, and both types are different: shared_ptr<Parent> vs. shared_ptr<Child>. Therefore, the compiler cannot figure out which type you mean by Child::ptr_t in main.

As you pointed out, you have to fix this manually using a typedef in Child (making your Pointable base class useless, though).

class Child : public Parent,
              public Pointable<Child> {
public:
    typedef Pointable<Child>::ptr_t ptr_t;