According to ISO standard you have explicitly qualify
the "Base" members:
template <typename _type>
class AnotherClass : public MyClass<_type>
{
public:
void print( void ) const
{
// THE ERROR IS ON THIS LINE
std::cerr << "m_value = " << MyClass<_type>::m_value << '\n';
}
};
gcc 3.4 is mutch more closed to ISO.
Read carefully info about gcc 3.4:
http://gcc.gnu.org/gcc-3.4/changes.html
----
Lev Assinovsky
Aelita Software Corporation
(now is a part of Quest Software)
O&S InTrust Framework Division, Team Leader
ICQ# 165072909
> -----Original Message-----
> From: gcc-help-owner@xxxxxxxxxxx [mailto:gcc-help-owner@xxxxxxxxxxx]On
> Behalf Of Laurent Marzullo
> Sent: Wednesday, June 09, 2004 2:08 PM
> To: gcc-help@xxxxxxxxxxx
> Subject: Inherite from template witth gcc 3.4
>
>
> Hello,
>
> Could someone explain why this do not compile anymore with GCC 3.4 ?
> (and compile with 3.2)
>
> #include <iostream>
>
> template <typename _type>
> class MyClass
> {
> public:
> _type m_value;
> };
>
> template <typename _type>
> class AnotherClass : public MyClass<_type>
> {
> public:
> void print( void ) const
> {
> // THE ERROR IS ON THIS LINE
> std::cerr << "m_value = " << m_value << '\n';
> }
> };
>
> int
> main( void )
> {
> AnotherClass<bool> a;
> a.print();
> return 0;
> }
>
>
>
> template.cpp: In member function `void AnotherClass<_type>::print()
> const':
> template.cpp:18: error: `m_value' undeclared (first use this function)
> template.cpp:18: error: (Each undeclared identifier is reported only
> once for each function it appears in.)
>
> I should specify, explicitly, MyClass<_type>::m_value for the program
> to compile. Why should I specify the base class for accessing this
> member ?
>
> Is there any way to go around this 'probleme' with gcc 3.4 ?
> Thanks
> Laurent Marzullo
>
>
>
