| re: problem with explicit template instantiation in Visual C++ 6.0 .
"Jonathan Turkanis" <technews@kangaroologic.com> wrote in message news:<bvsgor$vs42v$1@ID-216073.news.uni-berlin.de>...[color=blue]
> "C. Carbonera" <ccarbonera@msn.com> wrote in message
> news:dfb2f99f.0402041520.7e5f9fa5@posting.google.c om...[color=green]
> > /*
> > Hi,
> >
> > I have a problem with explicit instantiation of templates in Visual
> > C++ 6.0.
> > I have provided the source below. I have an example of a function
> > template that produces incorrect output in the code below.
> >
> > The function
> > template < typename T >
> > void DummyFunctionDoesNotWork(T &o);
> >
> > Produces incorrect output. The following is the expected output:
> > First Type
> > Second Type
> >
> > But, instead, I obtain the following is the output:
> > Second Type
> > Second Type
> >
> > I have written a second function
> >
> > template < typename T >
> > void DummyFunctionWorks(T &o);
> >
> > That produces the correct output.
> >
> > Could someone please test the code given below in Visual C++.NET
> > and let me know if it works?
> >
> > Thanks,
> > CCarbonera
> > */
> > #include <fstream.h>
> >
> > enum explicit_t
> > {first_t=1,
> > second_t=0};
> >
> > template <explicit_t t>
> > struct Dummy
> > {
> > const char* operator()(void){ return " UNDEFINED "; };
> > };
> >
> > template <>
> > const char* Dummy< first_t >::operator()(){ return " First Type "; }
> >
> > template <>
> > const char* Dummy<second_t>::operator()(){ return " Second Type "; }
> >
> > template < typename T >
> > void DummyFunctionWorks(T &o)
> > { cout << o() << endl; }
> >
> > template < typename T >
> > void DummyFunctionDoesNotWork()
> > { T o; cout << o() << endl; }
> >
> > void main(int argc, char* argv[])
> > {
> > cout<< "The following produces the correct output:" << endl;
> > Dummy<first_t> frst;
> > Dummy<second_t> scnd;
> > DummyFunctionWorks < Dummy< first_t > > ( frst );
> > DummyFunctionWorks < Dummy< second_t > > ( scnd );
> >
> > cout<< "The following produces the wrong output:" << endl;
> > DummyFunctionDoesNotWork < Dummy < first_t > > ( );
> > DummyFunctionDoesNotWork < Dummy < second_t > > ( );
> > }[/color]
>
> This is a very funny bug! Have you tried reversing the last two lines?
> I get:
>
> First Type
> First Type
>
> Both calls are resolved differently, juist because they are invoked in
> a different order.
>
> (By the way, you are really talking about explicit specializaion, not
> explicit instantiation. Also main returns int, ... )
>
> Jonathan[/color]
Jonathan,
Thank you for your feedback.
I have one comment. The problem occurs when the functions
void DummyFunctionDoesNotWork < Dummy < first_t > > ( ) and
void DummyFunctionDoesNotWork < Dummy < second_t > > ( ) are
instantiated;
hence, I labeled the problem as an eplicit instantiation.
There explicit specializations of the functions
const char* Dummy< first_t >::operator()() and
const char* Dummy< second_t >::operator()()
work well as the example below illustrates.
/*
Please, turn on and off the compiler directive '#define WORKS' below
to compare the results.
*/
#include <iostream>
using namespace std;
enum explicit_t
{
first_t=1,
second_t=0
};
template <explicit_t t>
struct Dummy{ virtual const char* operator()(void){ return NULL; };
};
template<>
const char* Dummy<first_t>::operator()(void) { return "First
Function"; }
template<>
const char* Dummy<second_t>::operator()(void) { return "Second
Function"; }
//#define WORKS
#ifdef WORKS
template <typename T>
void DummyFunction(T o = T())
{
cout << T()() << endl;
}
#else
template <typename T>
void DummyFunction( )
{
T o;
cout << o() << endl;
}
#endif
void main(void)
{
DummyFunction< Dummy<first_t> >();
DummyFunction< Dummy<second_t> >();
}
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