Weird compiler error and template inheritance

Question

Could someone explain me why this code:

class safe_bool_base
{ //13
    protected:

        typedef void (safe_bool_base::*bool_type)() const;

        void this_type_does_not_support_comparisons() const {} //18

        safe_bool_base() {}
        safe_bool_base(const safe_bool_base&) {}
        safe_bool_base& operator=(const safe_bool_base&) { return *this; }
        ~safe_bool_base() {}
};

template <typename T=void> class safe_bool : public safe_bool_base
{
    public:

        operator bool_type() const
        {
            return (static_cast<const T*>(this))->boolean_test() ? &safe_bool_base::this_type_does_not_support_comparisons : 0;
        }

    protected:

        ~safe_bool() {}
};

template <> class safe_bool<void> : public safe_bool_base
{
    public:

        operator bool_type() const
        {
            return (boolean_test() == true) ? &safe_bool_base::this_type_does_not_support_comparisons : 0; //46
        }

    protected:

        virtual bool boolean_test() const = 0;
        virtual ~safe_bool() {}
};

Produces the following compiler error ?

c:\project\include\safe_bool.hpp(46) : error C2248: 'safe_bool_base::this_type_does_not_support_comparisons' : cannot access protected member declared in class 'safe_bool_base'
c:\project\include\safe_bool.hpp(18) : see declaration of 'safe_bool_base::this_type_does_not_support_comparisons'
c:\project\include\safe_bool.hpp(13) : see declaration of 'safe_bool_base'

Since both safe_bool templates derive from safe_bool_base, I don't understand why one can't access a protected member of the base class.

Am I missing something ?

Answer 1

I don't think this is anything to do with templates. Your example code can be reduced to this, and it still gives the equivalent error:

class A
{
    protected:
        typedef void (A::*type)() const;
        void foo() const {}
};


class B : public A
{
    public:
        operator type() const
        {
            return &A::foo;
        }
};

I believe the issue is you can't return member-function pointers to protected members in the public interface. (Edit: not true...)

Answer 2

This should probably help (reproducible in a non template situation also)

struct A{
protected:
    void f(){}
};

struct B : A{
    void g(){&A::f;}        // error, due to Standard rule quoted below
};

int main(){
}

VS gives "'A::f' : cannot access protected member declared in class 'A'"

For the same code, Comeau gives

"ComeauTest.c", line 7: error: protected function "A::f" (declared at line 3) is not accessible through a "A" pointer or object void g(){&A::f;} ^

"ComeauTest.c", line 7: warning: expression has no effect void g(){&A::f;}

Here is the fixed code which achieves the desired intentions

struct A{
protected:
    void f(){}
};

struct B : A{
    void g(){&B::f;}        // works now
};

int main(){
}

So, why does the first code snippet not work?

This is because of the following rule in the C++ Standard03

11.5/1- "When a friend or a member function of a derived class references a protected nonstatic member function or protected nonstatic data member of a base class, an access check applies in addition to those described earlier in clause 11.102) Except when forming a pointer to member (5.3.1), the access must be through a pointer to, reference to, or object of the derived class itself (or any class derived from that class) (5.2.5). If the access is to form a pointer to member, the nested-name-specifier shall name the derived class (or any class derived from that class).

So change the return within operator functions as follows

return (boolean_test() == true) ? &safe_bool<void>::this_type_does_not_support_comparisons : 0; //46 

return (static_cast<const T*>(this))->boolean_test() ? &typename safe_bool<T>::this_type_does_not_support_comparisons : 0; 

EDIT 2: Please ignore my explanations. David is right. Here is what it boils down to.

struct A{
protected:
    int x;
};

struct B : A{
    void f();
};

struct C : B{};

struct D: A{            // not from 'C'
};

void B::f(){
    x = 2;         // it's own 'A' subobjects 'x'. Well-formed

    B b;
    b.x = 2;       // access through B, well-formed

    C c;
    c.x = 2;       // access in 'B' using 'C' which is derived from 'B', well-formed.

    D d;
    d.x = 2;       // ill-formed. 'B' and 'D' unrelated even though 'A' is a common base
}

int main(){} 

Answer 3

Chubsdad's answer clarifies your question of why there's an error for the template specialization.

Now the following C++ standard rule

14.7.2/11 The usual access checking rules do not apply to names used to specify explicit
instantiations. [Note: In particular, the template arguments and names used in the function
declarator (including parameter types, return types and exception specifications) may be
private types or objects which would normally not be accessible and the template may be a
member template or member function which would not normally be accessible. — endnote]

would explain why the generic template instantiation wouldn't throw an error. It will not throw even if you have private access specifier.