Re: Re: [boost] [serialization] export macros

From: Christoph Ludwig (cludwig_at_[hidden])
Date: 2004-08-14 13:28:07

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
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