Making a template parameter a friend?

Question

Example:

template
class Base {
public:
    Base();
    friend class T;
};

Now this doesn\'t work... Is there a way of doing

Answer 1

It is explicitly disallowed in the standard, even if some versions of VisualStudio do allow it.

C++ Standard 7.1.5.3 Elaborated type specifiers, paragraph 2

3.4.4 describes how name lookup proceeds for the identifier in an elaborated-type-specifier. If the identifier resolves to a class-name or enum-name, the elaborated-type-specifier introduces it into the declaration the same way a simple-type-specifier introduces its type-name. If the identifier resolves to a typedef-name or a template type-parameter, the elaborated-type-specifier is ill-formed. [Note: this implies that, within a class template with a template type-parameter T, the declaration friend class T; is ill-formed. ]

I recognize the code above as a pattern to seal (disallow the extension of) a class. There is another solution, that does not really block the extension but that will flag unadvertidly extending from the class. As seen in ADOBE Source Library:

namespace adobe { namespace implementation {
template 
class final
{
protected:
   final() {}
};
}}
#define ADOBE_FINAL( X ) private virtual adobe::implementation::final

with the usage:

class Sealed : ADOBE_FINAL( Sealed )
{//...
};

While it allows extension if you really force it:

class SealBreaker : public Sealed, ADOBE_FINAL( Sealed )
{
public:
   SealBreaker() : adobe::implementation::final(), Sealed() {}
};

It will restrict users from mistakenly do it.

EDIT:

The upcoming C++11 standard does allow you to befriend a type argument with a slightly different syntax:

template 
class A {
   // friend class T; // still incorrect: elaborate type specifier
   friend T;          // correct: simple specifier, note lack of "class"
};