Declare a template template as a friend

Question

For some reason, I would like to write code as follows:

template<class T> class C : public T { friend class T; }; 

I think the code is clear. I want to have a class template that defines a class derived from the class passed to it as a template parameter, and to make things a little more complex, I would like to define the base class as a friend of the derived one. The code looks fine with an MSVC compiler, but the GNU C ++ compiler complains a lot. What to do to have the desired functionality?

Answer 1

It is poorly formed and invalid C ++, although it works in MSVC. The C ++ 03 standard says this (7.1.5.3 §2):

3.4.4 describes how name lookups are performed for an identifier in a specified type specifier. If the identifier resolves to the class name or enum-name, the specified type specifier introduces it into the declare, just as the simple type specifier represents the type-name. If the identifier allows the name typedef or template type-parameter, the specified type-specifier is poorly formed. [Note: this means that in the template of the class with the template type-parameter T, declaration

  friend class T; 

badly formed. ] If the name search does not find an declaration for the name, the specified type specifier is poorly formed if it is not a simple class identifier for the class, in which case the identifier is as described in 3.3.1.

For the same reason, you cannot do things like friend class std::string; but you have to make friends with std::basic_string with the template options.

However, the new C ++ 11 specification allows you to use the new syntax to declare friends, which is simple (11.3 § 3 of N3242):

 friend <typename-specifier>; 

This new syntax allows you to do what you want (I don't know if MSVC still supports this):

 template<typename T> class C : public T { friend T; };