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:
>> 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> >
> 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
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