On Sat, Aug 14, 2004 at 10:57:36AM -0700, Paul Mensonides wrote:
I suppose another way to do it would be:
template<class T> struct id { typedef T type; };
template<class> struct pack; template<class T> struct pack<void (id<T>)> : id<T> { };
MY_MACRO(pack<void (id<void>)>::type) MY_MACRO(pack<void (id<MyTemplate<int, int>[10]>)>::type)
But an even better way is to do the unpacking in the macro and clean up the syntax:
struct pack;
template<class T> struct lock { typedef T type; };
template<class T> struct unpack : lock<T> { }; template<> struct unpack<pack (void)> : lock<void> { }; template<class T> struct unpack<pack (T)> : lock<T> { }; template<class T> struct unpack<pack (lock<T>)> : lock<T> { };
#define MY_MACRO(t) ... unpack<t>::type ...
MY_MACRO( int ) MY_MACRO( pack (std::pair<int, int>) ) MY_MACRO( pack (lock<int[10]>) )
Wow. I am always surprised what you can achieve with some template metaprogramming. And I consider your example particularly elegant because a user can ignore the machinery unless the naive approach fails. Very neat! Thanks for writing it down! Regards Christoph -- http://www.informatik.tu-darmstadt.de/TI/Mitarbeiter/cludwig.html LiDIA: http://www.informatik.tu-darmstadt.de/TI/LiDIA/Welcome.html