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

From: Eric Niebler (eric_at_[hidden])
Date: 2008-03-24 18:35:43

Larry Evans wrote:
> 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> >?

No, it's not. terminal<X> is a metafunction. terminal<X>::type is 
expr<tag::terminal, args0<X> >. IN ADDITION, terminal<X> is a grammar 
(see my other message). The one thing terminal<X> is NOT is an expression.

-- 
Eric Niebler
Boost Consulting
www.boost-consulting.com