[Bug c++/57713] New: Template functions see friendship as inherited

[Bug c++/57713] New: Template functions see friendship as inherited

[w.shane.grant at gmail dot com]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=57713

            Bug ID: 57713
           Summary: Template functions see friendship as inherited
           Product: gcc
           Version: 4.8.1
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: c++
          Assignee: unassigned at gcc dot gnu.org
          Reporter: w.shane.grant at gmail dot com

Template functions that are friends of a class bypass access control for
derived classes.

----------

template <class T> T&& declval();

template <class T>
auto foo() -> decltype(declval<T&>().bar()) {}

class Base
{
  private:
    template <class T>
    friend auto foo() -> decltype(declval<T&>().bar());

    void bar(){}
};

class Derived : public Base {};

int main()
{
  foo<Derived>();
}

-----------

bar should be private in Derived but the friend declaration in Base is allowing
foo to access it, preventing compilation from failing.  Here's another very
similar example:

---------

template <class T>
void foo(T&&t)
{
  t.bar();
}

class Base
{
  private:
    template <class T>
    friend void foo(T&&);

    void bar(){}
};

class Derived : public Base {};

int main()
{
  foo(Derived());
}

--------

Derived gets the ability to call bar() only inside of the template function
foo.

Happens on (4.9.0 20130625 (experimental)) and (4.8.1).

[Bug c++/57713] Template functions see friendship as inherited

[w.shane.grant at gmail dot com]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=57713

Shane <w.shane.grant at gmail dot com> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|UNCONFIRMED                 |RESOLVED
         Resolution|---                         |INVALID

--- Comment #1 from Shane <w.shane.grant at gmail dot com> ---
Marking invalid since I think this was actually a case of not interpreting
friendship correctly.  A simpler test case:

class Base {
private:             
  friend void foo(); 
  void bar();  
};

class Derived : public Base {};

void foo() { Derived().bar(); }

also compiles with no error because has access to Base (due to friendship) and
thus access to bar.