In accessible base class for return pointer type

[Steven T. Hatton]

Can somebody explain why making T a friend of either B or C will permit the
code to compile?

class A{ protected: A(){} };
class T;
class B: protected A {protected: B(){}/*friend class T;*/};
class C: public B {public : C(){}/*friend class T;*/};
struct T { A* newA() {return new C();}};
int main() {}

This is the error I get if I do not use one of the friend declarations:
In member function ?A* T::newA()?: error: ?A? is an inaccessible base of ?C?

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[Victor Bazarov]

Steven T. Hatton wrote:

Can somebody explain why making T a friend of either B or C will
permit the code to compile?

class A{ protected: A(){} };
class T;
class B: protected A {protected: B(){}/*friend class T;*/};
class C: public B {public : C(){}/*friend class T;*/};
struct T { A* newA() {return new C();}};
int main() {}

This is the error I get if I do not use one of the friend
declarations:
In member function ?A* T::newA()?: error: ?A? is an inaccessible base
of ?C?

Conversions from a pointer to derived to a pointer base require the
base to be accessible. If 'A' is a protected base of 'C' (through
'B'), then no outside class/function can convert a pointer to 'C' to
a pointer to 'A'. Since 'B' is a public base of 'C', anyone can
convert a 'C*' to a 'B*'.

What book on C++ are you reading that doesn't explain access specifiers
and their effect?
V
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[Steven T. Hatton]

Victor Bazarov wrote:

Steven T. Hatton wrote:

Can somebody explain why making T a friend of either B or C will
permit the code to compile?

class A{ protected: A(){} };
class T;
class B: protected A {protected: B(){}/*friend class T;*/};
class C: public B {public : C(){}/*friend class T;*/};
struct T { A* newA() {return new C();}};
int main() {}

This is the error I get if I do not use one of the friend
declarations:
In member function ?A* T::newA()?: error: ?A? is an inaccessible base
of ?C?

Conversions from a pointer to derived to a pointer base require the
base to be accessible. If 'A' is a protected base of 'C' (through
'B'), then no outside class/function can convert a pointer to 'C' to
a pointer to 'A'. Since 'B' is a public base of 'C', anyone can
convert a 'C*' to a 'B*'.

What book on C++ are you reading that doesn't explain access specifiers
and their effect?
V

I guess I could/should have looked it up. I just wanted to see what others
had to say about it. After your suggestion that I access a reference, I
did take a look at TC++PL(SE) §15.3.2. Basically what it says, and what
you said are the supposition I drew from the example. Part of what was
unclear in my mind is what scope the conversion takes place in. The
original code was calling the counterpart of C::newA (above) from within
the counterpart of A. It took some time to untangle what was happening
because I had never encountered that particular situation, that I can
remember.
--
NOUN:1. Money or property bequeathed to another by will. 2. Something handed
down from an ancestor or a predecessor or from the past: a legacy of
religious freedom. ETYMOLOGY: MidE legacie, office of a deputy, from OF,
from ML legatia, from L legare, to depute, bequeath. www.bartleby.com/61/