template typedef as return type

[Robert A. T. Kaldy]

Hello,

consider a class with a member function:

template<class schema>
class A
{
schema::t func();
};

template<class schema>
schema::t A<schema>::func()
{ .... }

The 't' is a type, which will be defined via typedef later, in the schema
class.
It takes the error
'f': unable to resolve function overload
(I compiled it with VC++ 5.0)
Putting "typename schema" or "typename schema::t" in the declaration of A
didn't solved it.
Any ideas?

Robert-Antonio

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[Michael Kochetkov]

"Robert A. T. Kaldy" <ka****@matfyz.cz> wrote in message
news:Pi**************************************@arta x.karlin.mff.cuni.cz...

Hello,

consider a class with a member function:

template<class schema>
class A
{
schema::t func();
};

template<class schema>
schema::t A<schema>::func()
{ .... }

The 't' is a type, which will be defined via typedef later, in the schema
class. No, it is not. It is not a type but the object name. If you need to make a
compiler to treat it as a type you shall use "typename" qualifier.
It takes the error
'f': unable to resolve function overload
(I compiled it with VC++ 5.0)

Just a casual remark: I suspect you are going to have a lot of C++ language
problems with this pretty old compiler.

--
With regards,
Michael Kochetkov.