Declaring friend class template via wrapper

Question

I've seen the following pre-C++11 code, used as a trick to declare class template friends (which in C++11 can simply be done with friend T;)

template <typename T>
struct Wrapper
{
    typedef T type;
};

template <typename T> 
class Foo
{
    friend class Wrapper<T>::type; // effectively makes T a friend
};

struct Test{};

int main()
{
    Foo<Test> foo;
}

The code compiles fine on g++ (4.9/5.1/6), but fails under clang++ (3.5/3.6/3.7) with the error

error: elaborated type refers to a typedef

friend class Wrapper::type;

Is the code above standard compliant, i.e. valid or not?

Related: http://stackoverflow.com/questions/21952658/type-dependent-nested-name-specifier-in-elaborated-type-specifier — David G, May 21 '15 at 18:33
or http://stackoverflow.com/questions/14623338/elaborated-type-refers-to-typedef-error-when-trying-to-befriend-a-typedef — Anton Savin, May 21 '15 at 18:48

score 1 · Accepted Answer · answered May 21 '15 at 19:04

1

§7.1.6.3/2:

If the identifier resolves to a typedef-name or the simple-template-id resolves to an alias template specialization, the elaborated-type-specifier is ill-formed.

answered May 21 '15 at 19:04

Columbo

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score 0 · Answer 2 · answered May 21 '15 at 19:02

0

It's not compliant. The grammar rules for friend in [class.friend]/3 are:

A friend declaration that does not declare a function shall have one of the following forms:
friend elaborated-type-specifier ;
friend simple-type-specifier ;
friend typename-specifier ;

class Wrapper<T>::type is none of those specifier types. It's not an elaborated-type-specifier because Wrapper<T>::type isn't an identifier or a class-name, and obviously isn't one of the other two either. What you're looking for is simply:

friend typename Wrapper<T>::type;

answered May 21 '15 at 19:02

Barry

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`friend class-key nested-name-specifier identifier ;` is allowed by 7.1.6.3 – Columbo May 21 '15 at 19:03
@Columbo So if `Wrapper::type` was some actual type, rather than an alias, you could do `friend class Wrapper::type`? Why allow a type but not an alias? – Barry May 21 '15 at 19:07
Yes, that is valid. Not sure about the second part, but I'll look for a DR. – Columbo May 21 '15 at 19:12

David G · Answer 3 · 2015-05-21T19:08:11.200

[dcl.typedef]/p8:

[ Note: A typedef-name that names a class type, or a cv-qualified version thereof, is also a class-name (9.1)

If a typedef-name is used to identify the subject of an elaborated-type-specifier (7.1.6.3), a class definition (Clause 9), a constructor declaration (12.1), or a destructor declaration (12.4), the program is ill-formed. — end note ] [Example:
struct S {
    S();
   ~S();
};

typedef struct S T;
S a = T(); // OK
struct T * p; // error
— end example ]

The code should fail at template instantiation time, which it does so correctly in Clang.

Using typename in place of struct allows the code to pass in both compilers.

Well, that's a note; Not normative. – Columbo May 21 '15 at 19:05 — Columbo, May 21 '15 at 19:05

Declaring friend class template via wrapper

3 Answers3