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comp.lang.c++

Templates and inheritance

Jim West

11/8/2008 6:06:00 PM

Can someone please explain to me why the following compiles:

class A {
public:
int d;
};

class B : public A {
public:
int e;
void f() { d = e; };
};



but if I make A and B templates it gives an error:

template <typename T>
class A {
public:
T d;
};

template <typename T>
class B : public A<T> {
public:
int e;
void f() { d = e; }; // error: identifier "d" is undefined
};


This occurs with GNU G++ and Intel ICC. Obviously I am missing a subtlety
of templates and/or inheritance. (I figured out that changing the line in
question to

void f() { A<T>::d = e; };

corrects the error, but I want understand why it is an error in the first
place.)
2 Answers

Salt_Peter

11/8/2008 9:38:00 PM

0

On Nov 8, 1:06 pm, Jim West <eggplantpa...@yahoo.com> wrote:
> Can someone please explain to me why the following compiles:
>
> class A {
> public:
> int d;
>
> };
>
> class B : public A {
> public:
> int e;
> void f() { d = e; };
>
> };
>
> but if I make A and B templates it gives an error:
>
> template <typename T>
> class A {
> public:
> T d;
>
> };
>
> template <typename T>
> class B : public A<T> {
> public:
> int e;
> void f() { d = e; }; // error: identifier "d" is undefined
>
> };
>
> This occurs with GNU G++ and Intel ICC. Obviously I am missing a subtlety
> of templates and/or inheritance. (I figured out that changing the line in
> question to
>
> void f() { A<T>::d = e; };
>
> corrects the error, but I want understand why it is an error in the first
> place.)

During two-phase name lookup, base classes depending on a template
argument are not considered. Recall that a member function has a
hidden this parameter.
So replace B::f() with a free function instead:

template< typename T >
void f( B<T>& r_b )
{
r_b.d = r_b.e;
}

Hendrik Schober

11/9/2008 12:22:00 AM

0

Jim West wrote:
> Can someone please explain to me why the following compiles:
>
> class A {
> public:
> int d;
> };
>
> class B : public A {
> public:
> int e;
> void f() { d = e; };
> };
>
>
>
> but if I make A and B templates it gives an error:
>
> template <typename T>
> class A {
> public:
> T d;
> };
>
> template <typename T>
> class B : public A<T> {
> public:
> int e;
> void f() { d = e; }; // error: identifier "d" is undefined
> };

When the compiler encounters 'B' for the first time, it hasn't
been instantiated, so it doesn't know what 'A' looks like.
(Remember that 'A' could be specialized for some specific 'T'
up to the point where 'B' is instantiated for the first time.)
If it doesn't know 'A', it doesn't know 'A<T>::d'. If you spell
'A<A>::d' out explicitly, it becomes a so called dependent type
(which means it depends on some template parameter which isn't
known when the template is parsed for the first time) and its
lookup is deferred until 'B' is actually instantiated.

> This occurs with GNU G++ and Intel ICC. Obviously I am missing a subtlety
> of templates and/or inheritance. (I figured out that changing the line in
> question to
>
> void f() { A<T>::d = e; };
>
> corrects the error, but I want understand why it is an error in the first
> place.)

HTH,

Schobi