Base class attribute visibility

[Arne Schmitz]

Why does the following code not compile on g++-4.0? g++-3.3 does work.

----------
template<typename Scalar, int N>
struct FooData
{
int i;
};

template<typename Scalar, int N>
class Foo
: public FooData<Scalar,N>
{
public:
void test()
{
i = 0;
}
};
----------
$ g++-4.0 -c template.cc -o template.o
template.cc: In member function 'void Foo<Scalar, N>::test()':
template.cc:14: error: 'i' was not declared in this scope
----------

If I add a FooData<Scalar,N>:: in front of the i, it works with g++-4.0. But
i should be visible in Foo without this, right?

Arne

--
[--- PGP key FD05BED7 --- http://www.root42.de/ ---]

[Rolf Magnus]

Arne Schmitz wrote:

Why does the following code not compile on g++-4.0? g++-3.3 does work.

----------
template<typename Scalar, int N>
struct FooData
{
int i;
};

template<typename Scalar, int N>
class Foo
: public FooData<Scalar,N>
{
public:
void test()
{
i = 0;
}
};
----------
$ g++-4.0 -c template.cc -o template.o
template.cc: In member function 'void Foo<Scalar, N>::test()':
template.cc:14: error: 'i' was not declared in this scope
----------

If I add a FooData<Scalar,N>:: in front of the i, it works with g++-4.0.
But i should be visible in Foo without this, right?

Try:

this->i = 0;

This is some obscure C++ rule that I never really understood. Maybe someone
else can explain the reasons for that. It has something to do with name
lookup in class templates.

[mlimber]

Rolf Magnus wrote:

Arne Schmitz wrote:

Why does the following code not compile on g++-4.0? g++-3.3 does work.

----------
template<typename Scalar, int N>
struct FooData
{
int i;
};

template<typename Scalar, int N>
class Foo
: public FooData<Scalar,N>
{
public:
void test()
{
i = 0;
}
};
----------
$ g++-4.0 -c template.cc -o template.o
template.cc: In member function 'void Foo<Scalar, N>::test()':
template.cc:14: error: 'i' was not declared in this scope
----------

If I add a FooData<Scalar,N>:: in front of the i, it works with g++-4.0.
But i should be visible in Foo without this, right?

Try:

this->i = 0;

This is some obscure C++ rule that I never really understood. Maybe someone
else can explain the reasons for that. It has something to do with name
lookup in class templates.

For the details, see this thread:

http://groups.google.com/group/comp....e537cee49e8dba

Especially the post by Hyman Rosen.

Cheers! --M

[Greg Comeau]

In article <3u************@news.dfncis.de>,
Arne Schmitz <ar**********@gmx.net> wrote:

Why does the following code not compile on g++-4.0? g++-3.3 does work.

----------
template<typename Scalar, int N>
struct FooData
{
int i;
};

template<typename Scalar, int N>
class Foo
: public FooData<Scalar,N>
{
public:
void test()
{
i = 0;
}
};
----------
$ g++-4.0 -c template.cc -o template.o
template.cc: In member function 'void Foo<Scalar, N>::test()':
template.cc:14: error: 'i' was not declared in this scope
----------

If I add a FooData<Scalar,N>:: in front of the i, it works with g++-4.0. But
i should be visible in Foo without this, right?

http://www.comeaucomputing.com/techt...membernotfound
--
Greg Comeau / Celebrating 20 years of Comeauity!
Comeau C/C++ ONLINE ==> http://www.comeaucomputing.com/tryitout
World Class Compilers: Breathtaking C++, Amazing C99, Fabulous C90.
Comeau C/C++ with Dinkumware's Libraries... Have you tried it?

[Arne Schmitz]

Greg Comeau wrote:

If I add a FooData<Scalar,N>:: in front of the i, it works with g++-4.0.
But i should be visible in Foo without this, right?

http://www.comeaucomputing.com/techt...membernotfound

Thanks, I only understand half of it, and cannot imagine an example where it
really makes sense to handle things like this, but at least I know what to
do.

Arne

--
[--- PGP key FD05BED7 --- http://www.root42.de/ ---]

[Arne Schmitz]

mlimber wrote:

For the details, see this thread:

http://groups.google.com/group/comp....e537cee49e8dba
Especially the post by Hyman Rosen.

Also interesting, but can you give an example, where this lookup rule is
really useful?

Arne

--
[--- PGP key FD05BED7 --- http://www.root42.de/ ---]

[Greg Comeau]

In article <3u************@news.dfncis.de>,
Arne Schmitz <ar**********@gmx.net> wrote:

Greg Comeau wrote:

If I add a FooData<Scalar,N>:: in front of the i, it works with g++-4.0.
But i should be visible in Foo without this, right?

http://www.comeaucomputing.com/techt...membernotfound

Thanks, I only understand half of it, and cannot imagine an example where it
really makes sense to handle things like this, but at least I know what to
do.

One way to think of this is that the template is just a description
of an eventual class. What that actual class looks like
we don't know until all the implicit and explicit stuff is done.
Until that, it doesn't know which base members are really there or not.
Note that the extra prods to direct the compiler are not guarantees
(they could be "lies") but there really is no other practical way to
accomplish the intent of melding the extremes involved.
--
Greg Comeau / Celebrating 20 years of Comeauity!
Comeau C/C++ ONLINE ==> http://www.comeaucomputing.com/tryitout
World Class Compilers: Breathtaking C++, Amazing C99, Fabulous C90.
Comeau C/C++ with Dinkumware's Libraries... Have you tried it?

[Arne Schmitz]

Greg Comeau wrote:

One way to think of this is that the template is just a description
of an eventual class. What that actual class looks like
we don't know until all the implicit and explicit stuff is done.
Until that, it doesn't know which base members are really there or not.
Note that the extra prods to direct the compiler are not guarantees
(they could be "lies") but there really is no other practical way to
accomplish the intent of melding the extremes involved.

Ok, thank you!

Arne

--
[--- PGP key FD05BED7 --- http://www.root42.de/ ---]

[red floyd]

Arne Schmitz wrote:

Greg Comeau wrote:

If I add a FooData<Scalar,N>:: in front of the i, it works with g++-4.0.
But i should be visible in Foo without this, right?

http://www.comeaucomputing.com/techt...membernotfound

Thanks, I only understand half of it, and cannot imagine an example where it
really makes sense to handle things like this, but at least I know what to
do.

consider this subcase of your example:

template<typename Scalar, int N>
struct FooData
{
int i;
};

template<>
struct FooData<void*, 0>
{
int haha_i_is_not_declared;
};

template<typename Scalar, int N>
class Foo
: public FooData<Scalar,N>
{
public:
void test()
{
i = 0;
}
};

[Clark S. Cox III]

On 2005-11-17 10:27:14 -0500, Rolf Magnus <ra******@t-online.de> said:

Arne Schmitz wrote:

Why does the following code not compile on g++-4.0? g++-3.3 does work.

----------
template<typename Scalar, int N>
struct FooData
{
int i;
};

template<typename Scalar, int N>
class Foo
: public FooData<Scalar,N>
{
public:
void test()
{
i = 0;
}
};
----------
$ g++-4.0 -c template.cc -o template.o
template.cc: In member function 'void Foo<Scalar, N>::test()':
template.cc:14: error: 'i' was not declared in this scope
----------

If I add a FooData<Scalar,N>:: in front of the i, it works with g++-4.0.
But i should be visible in Foo without this, right?

Try:

this->i = 0;

This is some obscure C++ rule that I never really understood. Maybe someone
else can explain the reasons for that. It has something to do with name
lookup in class templates.

The compiler can't know for sure that you won't add some specialization
of the parent type that either doesn't have a member variable named
"i", or has a type named "i":

template<int N>
class FooData<char, N>
{
//Whoops, there's no member variable 'i'
};

template<int N>
class FooData<unsigned char, N>
{
//Whoops, there a type named 'i'
typedef int i;
};

Now think about what that would mean for the subclass.

--
Clark S. Cox, III
cl*******@gmail.com