template<class T> void ReportType()
{
 if(typeid(T) == typeid(string))
	 cout<<"It''s a string! "<<endl;
 else if(typeid(T) == typeid(int))
	 cout<<"It''s a int! "<<endl;
}

int main()
{
ReportType<int>();
ReportType<string>();
return 1;
}

#include <iostream>template<class T> void ReportType(){	if(typeid(T) == typeid(std::string)) std::cout << "It's a string!" << std::endl;	else if(typeid(T) == typeid(int)) std::cout << "It's a int!" << std::endl;}int main(){	ReportType<int>();	ReportType<std::string>();	return 0;}

quote:Original post by noVum
Visual Studio .NET 2003

program:

#include <iostream>template<class T> void ReportType(){	if(typeid(T) == typeid(std::string)) std::cout << "It''s a string!" << std::endl;	else if(typeid(T) == typeid(int)) std::cout << "It''s a int!" << std::endl;}int main(){	ReportType<int>();	ReportType<std::string>();	return 0;}

output:

It''s a int!
It''s a string!

[edited by - novum on July 8, 2003 12:21:08 PM]

#include <iostream>#include <string>using namespace std;

quote:Original post by Fruny
Visual Studio 6 has problems with template functions which do not take an argument of their templated type.

[ Start Here ! | How To Ask Smart Questions | Recommended C++ Books | C++ FAQ Lite | Function Ptrs | CppTips Archive ]
[ Header Files | File Format Docs | LNK2001 | C++ STL Doc | STLPort | Free C++ IDE | Boost C++ Lib | MSVC6 Lib Fixes ]

quote:Original post by Burning_Ice
well it shouldn't matter much, since when will you use a function such as the above with a template param but return type void and without arguments?

the real power of templates lies in generic programming, by letting the compiler deduce the types of the function.
take as an example this function:

template < class T > T mult ( const T& t1, const T& t2 ){	return t1 * t2;}

now the beauty is, that you can use this function with any two values of a type that supports copying and operator*
thus
cout

 Function<typename>();However would the broblem be fixed if I would't use <typename> whit call? like:template<class T> T tempfn(T t,int pos){// never use treturn *(T*)stack[pos];}int i; // useless var (my way to fix...)int rval = tempfn(i,0);now the type can be know but it's kinda dirty coding ain't it?