CRTP can't see structs define in derived class [duplicate] - c++

Question

I read the Wikipedia article about the curiously recurring template pattern in C++ for doing static (read: compile-time) polymorphism. I wanted to generalize it so that I could change the return types of the functions based on the derived type. (This seems like it should be possible since the base type knows the derived type from the template parameter). Unfortunately, the following code won't compile using MSVC 2010 (I don't have easy access to gcc right now so I haven't tried it yet). Anyone know why?
template <typename derived_t>
class base {
public:
typedef typename derived_t::value_type value_type;
value_type foo() {
return static_cast<derived_t*>(this)->foo();
}
};
template <typename T>
class derived : public base<derived<T> > {
public:
typedef T value_type;
value_type foo() {
return T(); //return some T object (assumes T is default constructable)
}
};
int main() {
derived<int> a;
}
BTW, I have a work-around using extra template parameters, but I don't like it---it will get very verbose when passing many types up the inheritance chain.
template <typename derived_t, typename value_type>
class base { ... };
template <typename T>
class derived : public base<derived<T>,T> { ... };
EDIT:
The error message that MSVC 2010 gives in this situation is error C2039: 'value_type' : is not a member of 'derived<T>'
g++ 4.1.2 (via codepad.org) says error: no type named 'value_type' in 'class derived<int>'

Answer 1

derived is incomplete when you use it as a template argument to base in its base classes list.
A common workaround is to use a traits class template. Here's your example, traitsified. This shows how you can use both types and functions from the derived class through the traits.
// Declare a base_traits traits class template:
template <typename derived_t>
struct base_traits;
// Define the base class that uses the traits:
template <typename derived_t>
struct base {
typedef typename base_traits<derived_t>::value_type value_type;
value_type base_foo() {
return base_traits<derived_t>::call_foo(static_cast<derived_t*>(this));
}
};
// Define the derived class; it can use the traits too:
template <typename T>
struct derived : base<derived<T> > {
typedef typename base_traits<derived>::value_type value_type;
value_type derived_foo() {
return value_type();
}
};
// Declare and define a base_traits specialization for derived:
template <typename T>
struct base_traits<derived<T> > {
typedef T value_type;
static value_type call_foo(derived<T>* x) {
return x->derived_foo();
}
};
You just need to specialize base_traits for any types that you use for the template argument derived_t of base and make sure that each specialization provides all of the members that base requires.

Answer 2

One small drawback of using traits is that you have to declare one for each derived class. You can write a less verbose and redondant workaround like this :
template <template <typename> class Derived, typename T>
class base {
public:
typedef T value_type;
value_type foo() {
return static_cast<Derived<T>*>(this)->foo();
}
};
template <typename T>
class Derived : public base<Derived, T> {
public:
typedef T value_type;
value_type foo() {
return T(); //return some T object (assumes T is default constructable)
}
};
int main() {
Derived<int> a;
}

Answer 3

In C++14 you could remove the typedef and use function auto return type deduction:
template <typename derived_t>
class base {
public:
auto foo() {
return static_cast<derived_t*>(this)->foo();
}
};
This works because the deduction of the return type of base::foo is delayed until derived_t is complete.

Answer 4

An alternative to type traits that requires less boilerplate is to nest your derived class inside a wrapper class that holds your typedefs (or using's) and pass the wrapper as a template argument to your base class.
template <typename Outer>
struct base {
using derived = typename Outer::derived;
using value_type = typename Outer::value_type;
value_type base_func(int x) {
return static_cast<derived *>(this)->derived_func(x);
}
};
// outer holds our typedefs, derived does the rest
template <typename T>
struct outer {
using value_type = T;
struct derived : public base<outer> { // outer is now complete
value_type derived_func(int x) { return 5 * x; }
};
};
// If you want you can give it a better name
template <typename T>
using NicerName = typename outer<T>::derived;
int main() {
NicerName<long long> obj;
return obj.base_func(5);
}

Answer 5

I know that this is basically the workaround you found and don't like, but I wanted to document it and also to say that it is basically the current solution to this problem.
I have been looking for a way to do this for a while and never found a good solution.
The fact that it is not possible is the reason why ultimately, things like boost::iterator_facade<Self, different_type, value_type, ...> need many parameters.
Of course we would like something something like this to work:
template<class CRTP>
struct incrementable{
void operator++(){static_cast<CRTP&>(*this).increment();}
using ptr_type = typename CRTP::value_type*; // doesn't work, A is incomplete
};
template<class T>
struct A : incrementable<A<T>>{
void increment(){}
using value_type = T;
value_type f() const{return value_type{};}
};
int main(){A<double> a; ++a;}
If this was possible, all the traits of the derived class could be passed implicitly ot the base class. The idiom I found to get the same effect is to pass the traits to the base class entirely.
template<class CRTP, class ValueType>
struct incrementable{
void operator++(){static_cast<CRTP&>(*this).increment();}
using value_type = ValueType;
using ptr_type = value_type*;
};
template<class T>
struct A : incrementable<A<T>, T>{
void increment(){}
typename A::value_type f() const{return typename A::value_type{};}
// using value_type = typename A::value_type;
// value_type f() const{return value_type{};}
};
int main(){A<double> a; ++a;}
https://godbolt.org/z/2G4w7d
The drawback is that the trait in the derived class has to be accessed with a qualified typename or reenabled by using.