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James Ying
Guest
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 Posted: Tue Oct 28, 2003 5:55 pm Post subject: Type names defined in base template class not visible in der |
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Following is the complete code demonstrating the issue:
=====================
| Quote: | cat tt.cc
#include <iostream |
template
class Base {
public:
typedef T difference_type;
};
template <class T>
class Derived : public Base<T> {
public:
void test() {
difference_type t = 10;
std::cout << "t is " << t << std::endl;
}
};
int main()
{
Derived
d.test();
}
========== end of code ======================
This code compiles without warning in g++2.96, MIPSpro C++ Compilers:
Version 7.3.1.1m, Sun WorkShop 6 update 2 C++ 5.3 Patch 111685-10.
However g++3.2 gives warning: implicit typename is deprecated, please
see the documentation for details.
My question: is the warning necessary according to the C++ standard?
While I haven't read the standard carefully enough to answer the
question definitely, the C++PL book gives a similar code in the class
of Checked Iterator (pp. 562-563, 3rd editon):
template<class Cont, class Iter = typename Cont::iterator>
class Check_iter : public iterator_traits<Iter> {
// ....
reference_type operator*() {
// ...
Note the reference_type (rather, reference) is type-defined in the
base template iterator_traits<Iter>, and used directly in the derived
class without prefixing "typename".
One obvious approach to get rid of the g++ warning is to add the
following code in the derived class:
typedef typename Base<T>::difference_type difference_type;
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Andrey Tarasevich
Guest
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| Posted: Tue Oct 28, 2003 6:26 pm Post subject: Re: Type names defined in base template class not visible in |
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James Ying wrote:
| Quote: | Following is the complete code demonstrating the issue:
=====================
cat tt.cc
#include <iostream
template
class Base {
public:
typedef T difference_type;
};
template
class Derived : public Base
public:
void test() {
difference_type t = 10;
std::cout << "t is " << t << std::endl;
}
};
int main()
{
Derived
d.test();
}
========== end of code ======================
This code compiles without warning in g++2.96, MIPSpro C++ Compilers:
Version 7.3.1.1m, Sun WorkShop 6 update 2 C++ 5.3 Patch 111685-10.
However g++3.2 gives warning: implicit typename is deprecated, please
see the documentation for details.
My question: is the warning necessary according to the C++ standard?
While I haven't read the standard carefully enough to answer the
question definitely, the C++PL book gives a similar code in the class
of Checked Iterator (pp. 562-563, 3rd editon):
|
I'd say that it should be an error, not just a warning. According to the
definition given in 14.6.2.1/1, name 'difference_type' used in the
definition of method 'test()' is not a dependent name. For this reason,
as 14.6/8 says, this name should be looked up is accordance with usual
name lookup rules at the point of template definition. Since class
template B<T> does not participate in this name lookup process (see
14.6.2/3), the name 'difference_type' will not be found and the code is
ill-formed. That's actually how Comeau's compiler treats this code - it
issues an error
"ComeauTest.c", line 13: error: identifier "difference_type" is undefined
difference_type t = 10;
^
| Quote: | template<class Cont, class Iter = typename Cont::iterator
class Check_iter : public iterator_traits
// ....
reference_type operator*() {
// ...
Note the reference_type (rather, reference) is type-defined in the
base template iterator_traits<Iter>, and used directly in the derived
class without prefixing "typename".
|
Well, the code in the book can be incomplete or outdated.
| Quote: | One obvious approach to get rid of the g++ warning is to add the
following code in the derived class:
typedef typename Base<T>::difference_type difference_type;
|
That would be the right thing to do.
--
Best regards,
Andrey Tarasevich
Brainbench C and C++ Programming MVP
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James Ying
Guest
|
| Posted: Tue Oct 28, 2003 11:36 pm Post subject: Re: Type names defined in base template class not visible in |
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|
Thank you for pointing to the right sections in the standard, and I
agree with you. The standard is pretty clean on this, although it
makes life a bit mroe difficult -- the supposed trick presented in
TC+PL about checked iterator will not do the trick to save a few
typings.
Andrey Tarasevich <andreytarasevich (AT) hotmail (DOT) com> wrote
| Quote: | James Ying wrote:
Following is the complete code demonstrating the issue:
=====================
cat tt.cc
#include <iostream
template
class Base {
public:
typedef T difference_type;
};
template
class Derived : public Base
public:
void test() {
difference_type t = 10;
std::cout << "t is " << t << std::endl;
}
};
int main()
{
Derived
d.test();
}
========== end of code ======================
This code compiles without warning in g++2.96, MIPSpro C++ Compilers:
Version 7.3.1.1m, Sun WorkShop 6 update 2 C++ 5.3 Patch 111685-10.
However g++3.2 gives warning: implicit typename is deprecated, please
see the documentation for details.
My question: is the warning necessary according to the C++ standard?
While I haven't read the standard carefully enough to answer the
question definitely, the C++PL book gives a similar code in the class
of Checked Iterator (pp. 562-563, 3rd editon):
I'd say that it should be an error, not just a warning. According to the
definition given in 14.6.2.1/1, name 'difference_type' used in the
definition of method 'test()' is not a dependent name. For this reason,
as 14.6/8 says, this name should be looked up is accordance with usual
name lookup rules at the point of template definition. Since class
template B<T> does not participate in this name lookup process (see
14.6.2/3), the name 'difference_type' will not be found and the code is
ill-formed. That's actually how Comeau's compiler treats this code - it
issues an error
"ComeauTest.c", line 13: error: identifier "difference_type" is undefined
difference_type t = 10;
^
template<class Cont, class Iter = typename Cont::iterator
class Check_iter : public iterator_traits
// ....
reference_type operator*() {
// ...
Note the reference_type (rather, reference) is type-defined in the
base template iterator_traits<Iter>, and used directly in the derived
class without prefixing "typename".
Well, the code in the book can be incomplete or outdated.
One obvious approach to get rid of the g++ warning is to add the
following code in the derived class:
typedef typename Base<T>::difference_type difference_type;
That would be the right thing to do.
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