CODEWITHSUNDEEP
c++templates
Raman Gupta
Raman Gupta

Reputation: 169

Please explain the meaning of this C++ template code

template<typename HELPER_>
class A{
     public:

        typedef HELPER_ HELPER ;
        typedef typename HELPER::TRAITS_  TRAITS_2;
     // other stuff....
};

My question is that HELPER_ is a type , then HELPER will also be a type, then what does HELPER::TRAITS_ signify. similarly if HELPER is not the name of class, it is just an unknown type being specified. But in the above code it feels like HELPER is the name of the class and it is calling its public variable name TRAITS_.

I want to know what the above code means.

Upvotes: 1

Views: 155

Answers (4)

Mark B
Mark B

Reputation: 96241

The code is allowing you to specify customized traits in the template class by using a separate class to specify those traits. It uses typedef to help complete the abstraction. For example, you could set up a trait class so that A::TRAITS_ works out to be int like this (noting that type names in all caps with a trailing underscore are somewhat uncommon):

class MyTraits
{
public:
    typedef int TRAITS_;
};

template<typename HELPER_>
class A{
     public:

        typedef HELPER_ HELPER ;
        typedef typename HELPER::TRAITS_  TRAITS_2;
     // other stuff....
};

// A<MyTraits>::TRAITS_2 is now `int`.

Upvotes: 1

Xeo
Xeo

Reputation: 131789

TRAITS_ is a nested name inside whatever HELPER_ happens to be. Since typename is put in front of this nested name, it's expected to signify a type.

Example:

#include <string>
#include <iostream>

template<class T>
struct X{
  typedef typename T::some_type some_type;
  X(some_type const& p)
    : m(p) {}

private:
  some_type m;
};

struct A{
  typedef int some_type;
};

struct B{
  typedef std::string some_type;
};

int main(){
  X<A> x1(5);
  X<B> x2("hi");
  std::cout << "X<A>::m = " << x1.m << ", X<B>::m = " << x2.m "\n";
}

Output:

X<A>::m = 5, X<B>::m = hi

Upvotes: 0

Drew Dormann
Drew Dormann

Reputation: 63765

This code is assuming that for any HELPER_ template parameter used, HELPER_::TRAITS_ exists and is a typename.

What the code is actually doing:

For a type X, this code is establishing that A<X>::TRAITS_2 is defined as the same type as X::TRAITS_

Upvotes: 1

zch
zch

Reputation: 15278

typedef HELPER_ HELPER ;

Aliases HELPER_ as HELPER in class namespace, so A<B>::HELPER becomes equivalent to B. It can useful in template code.

typedef typename HELPER::TRAITS_  TRAITS_2;

Aliases HELPER::TRAITS_ as TRAITS_2 in class namespace, so A<B>::TRAITS_2 becomes equivalent to B::TRAITS_. typename is necessary, becouse compiler doesn't know that HELPER::TRAITS_ is a type without definition of HELPER.

Upvotes: 1

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