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Home/ Questions/Q 6707121
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Editorial Team
  • 0
Editorial Team
Asked: May 26, 20262026-05-26T07:36:06+00:00 2026-05-26T07:36:06+00:00

Trying to make a class friends with an extern C function, this code works:

  • 0

Trying to make a class friends with an extern "C" function, this code works:

#include <iostream>

extern "C" {
  void foo();
}

namespace {
  struct bar {
    // without :: this refuses to compile
    friend void ::foo();
    bar() : v(666) {}
  private:
    int v;
  } inst;
}

int main() {
  foo();
}

extern "C" {
  void foo() {
    std::cout << inst.v << std::endl;
  }
}

But I was very surprised to find that with g++ 4.6.1 and 4.4.4 I have to explicitly write :: in friend void ::foo(); otherwise the friendship doesn’t work. This :: is only needed when it’s extern "C" though.

  1. Is this a compiler bug/problem? I wasn’t expecting that behaviour.
  2. If it isn’t a bug why is this required, but only when it’s extern "C" and not without it? What about the name lookup rules changes that makes this necessary?

I’m stumped. There is probably some rule for this that I can’t find.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T07:36:07+00:00Added an answer on May 26, 2026 at 7:36 am

    [n3290: 7.3.1.2/3]: Every name first declared in a namespace is a
    member of that namespace. If a friend declaration in a non-local class
    first declares a class or function the friend class or function is a
    member of the innermost enclosing namespace. The name of the friend is
    not found by unqualified lookup (3.4.1) or by qualified lookup (3.4.3)
    95) this implies that the name of the class or function is
    unqualified. until a matching declaration is provided in that
    namespace scope (either before or after the class definition granting
    friendship). If a friend function is called, its name may be found by
    the name lookup that considers functions from namespaces and classes
    associated with the types of the function arguments (3.4.2). If the
    name in a friend declaration is neither qualified nor a template-id
    and the declaration is a function or an elaborated-type-specifier, the
    lookup to determine whether the entity has been previously declared
    shall not consider any scopes outside the innermost enclosing
    namespace.
    [..]

    The innermost enclosing namespace is the anonymous one, and you didn’t qualify the function name, so the name is not found.

    The namespace need not be anonymous, either.

    Note that the extern "C" in the question is a red herring, as the following also fails for the same reason:

    void foo();
    
    namespace {
    struct T {
       friend void foo();
    
       private: void bar() { cout << "!"; }
    } t;
    }
    
    void foo() { t.bar(); }
    
    int main() {
       foo();
    }
    
    /*
    In function 'void foo()':
    Line 7: error: 'void<unnamed>::T::bar()' is private
    compilation terminated due to -Wfatal-errors.
    */
    

    [alternative testcase, adapted from your original code]

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