Nested protected class within namespace

[John S. Fine]

Why do you think NameSpace::convertInner is declared inside NameSpace?

You mention NameSpace::convertInner inside NameSpace in the friend 
declaration.  But I don't think declaring a function as a friend 
declares the function.

You define NameSpace::convertInner explicitly specifying by that 
definition that convertInner is in NameSpace, but that form cannot 
declare convertInner  in NameSpace.  (You can never use the form A::B 
when declaring B in A, only when defining or using B that was elsewhere 
declared in A).

Bill Spotz wrote:
>
> namespace NameSpace {
> class Outer {
> public:
>   class Inner {};
>   friend Inner * convertInner(int i);
> protected:
>   class InnerSetup : public Inner {};
> };
> }
>
> NameSpace::Outer::Inner * NameSpace::convertInner(int i) {
>   if (i == 0) return new NameSpace::Outer::Inner();
>   else        return new NameSpace::Outer::InnerSetup();
> }
> ----------
>
> It compiles fine with g++ version 4.0.1, but newer versions (4.2.1 and 
> 4.4.0) give me the following error:
>
> example.cpp:11: error: 'NameSpace::Outer::Inner* 
> NameSpace::convertInner(int)' should have been declared inside 
> 'NameSpace'
>
> It looks to me like the function IS declared inside 'NameSpace', so I 
> can't tell what the compiler is trying to tell me.
>
>
>