Re: Declare type in function template based on class t type

From:
"Jim Langston" <tazmaster@rocketmail.com>
Newsgroups:
comp.lang.c++
Date:
Fri, 22 Oct 2010 11:35:33 -0700
Message-ID:
<i9slhm$pj0$1@four.albasani.net>
"Alf P. Steinbach /Usenet" <alf.p.steinbach+usenet@gmail.com> wrote in
message news:i9olkb$lje$1@news.eternal-september.org...

* Jim Langston, on 21.10.2010 08:09:

Following does not compile for good reason(s).

template <class t>
GLvoid jglPrint( const jglFont& font, t text, jglVertex2& position ) {

// ...
if ( typeid( t ) == typeid( const char* ) )
std::basic_string <char> buffer( text );
else if ( typeid( t ) == typeid( const wchar_t * ) )
std::basic_string <wchar_t> buffer(text);

Basically I am accepting a few types for text anything that resembles
text.
char*, const char*, wchar_t*, std::basic_string<wchar_t>, etc..

I want to declare my buffer containing the same basic type as text (char,
wchar_t, u16, etc..). I can't quite figure out the syntax. I'm fairly
sure it
can be done but I hope in some elegant manner. Any suggestions?


A type traits class, that is, a class template with a typedef, specialized
for each relevant 't'.


Thanks, this seems to work.

   template <typename T> class Basetype {
   private:
      typedef void Type;
   };
   template<> class Basetype<const wchar_t*> {
   public:
      typedef wchar_t Type;
   };
   template<> class Basetype<const char*> {
   public:
      typedef char Type;
   };

std::basic_string<Basetype<T>::Type> buffer( text );

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