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

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> >?
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:

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