Re: [boost] lisp similarity? (was Re: [proto]purpose of binary_expr<T, L, R>::type ?

From: Larry Evans (cppljevans_at_[hidden])
Date: 2008-03-24 18:36:43

On 03/24/08 15:40, Eric Niebler wrote:
[snip]
> not. I'm pretty sure the use of ::type is wrong. Let me see if I can put 
> in words what an expression is, and what a grammar is, and what the 
> relationship is between the two. Maybe you can combine this with your 
> understanding of morphisms to formulate this.
> 
> An expression is one of:
> * expr<tag::terminal, args0<X> >
...
[snip]

But isn't terminal<X> the same as expr<tag::terminal,args0<X> >?
So, if terminal<X> is an expression (type), then the following:

   terminal<int> t_int={{1}};

should compile, but it doesn't.  Instead, the ::type has to be
appended to it.  That's the reason for the ::type in the
formula I gave in immediately previous post.  I think what
you're calling expression I'm calling expr_gram here:

http://archives.free.net.ph/message/20080323.205859.65a40e5e.en.html