Derived template classes

[Torsten Wiebesiek]

Hi folks,

I have have upgraded my compiler (gcc 3.3 -> 3.4) and the new version
performs two-stage (or dependent) name lookup:

http://gcc.gnu.org/onlinedocs/gcc/Name-lookup.html

Given a templated base class:

template <class T> struct Base
{
int f();
int i;
};

Following derived class doesn't compile anymore:

template <class T> struct Derived : public Base<T>
{
int g() { return i; }
int h() { return f();}
};


The correct implementation is:

template <class T> struct Derived : public Base<T>
{
int g() { return Base<T>::i; }
int h() { return Base<T>::f();}
};

This looks quite confusing, especially when you combine more members of the
base class in an expression.

The gcc docs suggest: "Alternativ ely, Base<T>::i might be brought into scope
by a using-declaration."

And here is my problem: what is the correct syntax of this using
declaration?

"using Base<T>::i;" doesn't work?

Regards,
Torsten

[Victor Bazarov]

Re: Derived template classes

Torsten Wiebesiek wrote:[color=blue]
> I have have upgraded my compiler (gcc 3.3 -> 3.4) and the new version
> performs two-stage (or dependent) name lookup:
>
> http://gcc.gnu.org/onlinedocs/gcc/Name-lookup.html
>
> Given a templated base class:
>
> template <class T> struct Base
> {
> int f();
> int i;
> };
>
> Following derived class doesn't compile anymore:
>
> template <class T> struct Derived : public Base<T>
> {
> int g() { return i; }
> int h() { return f();}
> };
>
>
> The correct implementation is:
>
> template <class T> struct Derived : public Base<T>
> {
> int g() { return Base<T>::i; }
> int h() { return Base<T>::f();}[/color]

Or

int g() { return this->i; }
int h() { return this->f(); }
[color=blue]
> };
>
> This looks quite confusing, especially when you combine more members of the
> base class in an expression.[/color]

Yes, it can look confusing.
[color=blue]
> The gcc docs suggest: "Alternativ ely, Base<T>::i might be brought into scope
> by a using-declaration."
>
> And here is my problem: what is the correct syntax of this using
> declaration?
>
> "using Base<T>::i;" doesn't work?[/color]

Is that a question? I mean, doesn't it? Please post the code in
its non-working shape. I got this, and it should be working fine:

template <class T> struct Base
{
int f();
int i;
};

template <class T> struct Derived : public Base<T>
{
using Base<T>::i;
using Base<T>::f;
int g() { return i; }
int h() { return f();}
};

int main()
{
Derived<int> di;
di.g();
di.h();
}

V

[mlimber]

Re: Derived template classes


Torsten Wiebesiek wrote:[color=blue]
> Hi folks,
>
> I have have upgraded my compiler (gcc 3.3 -> 3.4) and the new version
> performs two-stage (or dependent) name lookup:
>
> http://gcc.gnu.org/onlinedocs/gcc/Name-lookup.html
>
> Given a templated base class:
>
> template <class T> struct Base
> {
> int f();
> int i;
> };
>
> Following derived class doesn't compile anymore:
>
> template <class T> struct Derived : public Base<T>
> {
> int g() { return i; }
> int h() { return f();}
> };
>
>
> The correct implementation is:
>
> template <class T> struct Derived : public Base<T>
> {
> int g() { return Base<T>::i; }
> int h() { return Base<T>::f();}
> };
>
> This looks quite confusing, especially when you combine more members of the
> base class in an expression.
>
> The gcc docs suggest: "Alternativ ely, Base<T>::i might be brought into scope
> by a using-declaration."
>
> And here is my problem: what is the correct syntax of this using
> declaration?
>
> "using Base<T>::i;" doesn't work?[/color]

That should work at function or file scope, not class scope. Are you
sure it's in the right place? You could also do:

int g() { return this->i; }

Cheers! --M

[mlimber]

Re: Derived template classes

mlimber wrote:[color=blue]
> Torsten Wiebesiek wrote:[color=green]
> > Hi folks,
> >
> > I have have upgraded my compiler (gcc 3.3 -> 3.4) and the new version
> > performs two-stage (or dependent) name lookup:
> >
> > http://gcc.gnu.org/onlinedocs/gcc/Name-lookup.html
> >
> > Given a templated base class:
> >
> > template <class T> struct Base
> > {
> > int f();
> > int i;
> > };
> >
> > Following derived class doesn't compile anymore:
> >
> > template <class T> struct Derived : public Base<T>
> > {
> > int g() { return i; }
> > int h() { return f();}
> > };
> >
> >
> > The correct implementation is:
> >
> > template <class T> struct Derived : public Base<T>
> > {
> > int g() { return Base<T>::i; }
> > int h() { return Base<T>::f();}
> > };
> >
> > This looks quite confusing, especially when you combine more members of the
> > base class in an expression.
> >
> > The gcc docs suggest: "Alternativ ely, Base<T>::i might be brought into scope
> > by a using-declaration."
> >
> > And here is my problem: what is the correct syntax of this using
> > declaration?
> >
> > "using Base<T>::i;" doesn't work?[/color]
>
> That should work at function or file scope, not class scope. Are you
> sure it's in the right place? You could also do:
>
> int g() { return this->i; }
>
> Cheers! --M[/color]

Retraction: Victor is right. It should work at class scope. Post the
code.

Cheers! --M

[Torsten Wiebesiek]

Re: Derived template classes

Victor Bazarov wrote:[color=blue][color=green]
>> And here is my problem: what is the correct syntax of this using
>> declaration?
>>
>> "using Base<T>::i;" doesn't work?[/color]
>
> Is that a question? I mean, doesn't it? Please post the code in
> its non-working shape.[/color]

Sorry, you're right. The wrong code is in the function definition:

template <class T>
int Derived<T>::g()
{
using Base<T>::i;
return i;
}

template <class T>
int Derived<T>::h()
{
using Base<T>::f;
return f();
}
[color=blue]
> [snip]
>
> template <class T> struct Derived : public Base<T>
> {
> using Base<T>::i;
> using Base<T>::f;
> int g() { return i; }
> int h() { return f();}
> };
>
> [snip][/color]

I thought, the using declaration might work at function scope, but somehow
it isn't. At class scope it works fine. :-)

Thanks for your help, Victor.

Torsten