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From: Eric Niebler (eric_at_[hidden])
Date: 2008-03-03 20:29:37

Larry Evans wrote:
> On 03/03/08 15:51, Eric Niebler wrote:
> > Larry Evans wrote:
> >> or did you mean shift_right<L,R> is a grammar when L and R
> >> are grammars?
> >
> > That is also true. This is a short-hand, provided for convenience.
> >
> > shift_right<L,R>::type is *very* simple. It is expr<tag::shift_right,
> > args2<L, R> >. Always.
> Unexpected again. I jumped to the conclusion that expr was for
> expressions, not grammars.

A not unreasonable conclusion to jump to, considering that it's called
"expr" and not "expr_or_grammar" or "tagged_heterogeneous_container". I
feel the name "expr" is appropriate because it's more of an expression
than anything else due to its operator overloads that build other
expressions. The fact that Proto reuses this template as a grammar type
is done mainly for a the deep symmetry that results: any expression type
is also a grammar that matches itself -- matches<X,X> is trivially true,
regardless of how complicated X is. See below.

> So then is shift_right<L,R>::type, when L
> and R are grammars, the same as shift_right<L,R>?

Yes, they are equivalent.

> > It's just a 2-element container and a tag. If L
> > and R are expressions, it is an expression. If they're grammars, it is a
> > grammar, suitable for use as the second template parameter to
> > proto::matches. It is simple, IMO, and leads to a very straightforward
> > implementation of proto::matches<>. See, for instance, how the behavior
> > of proto::matches<> is specified in terms of expr<> instantiations:
> >
> <>
> Ah ha! The fog begins to lift. I should have read that. I
> apologize. Maybe if near the beginning of the documentation the
> concepts of expressions, grammars, and transforms were introduced, and
> then mention that a grammar was a type of expression... or at least
> was formed with the same template as an expression... Hmm.. Getting
> confused again. How can one distinguish a grammar from an expression
> if both are formed from the same template, i.e. the expr template?
> Maybe by starting from the base case, the terminal template?
> IOW, is the following:
> base_case)
> For any type T:
> terminal<T> is a grammar type.

Yes. Also insert here, for any grammar types T0,T1,... and transform type X:
          posit<T0> is a grammar type
          shift_right<T0,T1> is a grammar type
          not_<T0> is a grammar type
          or_<T0,T1,...> is a grammar type
          and_<T0,T1,...> is a grammar type
          if_<X,T0,T1> is a grammar type

> terminal<T>::type is an expression type.

Yes. It's also a grammar type. It matches itself (more or less -- see
the reference for proto::matches<>).

> inductive_case)
> If T0,T1,...,Tn are all grammar types, and Tag is an n-ary tag,
> then expr<Tag,T0,T2,...,Tn> is a grammar type.

Yes. (Minor correction, expr<Tag, argsN<T0,T1,...,Tn> > is a grammar
type.) So by induction, you can say posit< not_<X> >, which matches a
unary plus node whose operand does not match X. And posit< not_<X>
>::type means the same thing.

> If T0,T1,...,Tn are all expression types, and Tag is an n-ary tag,
> then expr<Tag,T0,T,...,Tn> is an expression type.

Yes. It is also a grammar type that matches itself.

> Otherwise, expr<Tag,T0,T1,...,Tn> is meaningless
> (a.k.a. undefined).

Well, for instance expr<tag::posit, args1<int> > is neither a grammar
nor an expression, that's true. But it's actually not meaningless or
undefined. Such types are often intermediate when applying some
transforms. But that's a separate issue.

> a correct description of how to distinguish a grammar type from an
> expression type?

As I hope I've shown, these two are not mutually exclusive, but I think
you've got the basic gist.

> [snip]
> >
> > What I can agree with is the following: the documentation currently
> > doesn't say what right_shift<L,R>::type means when L and R are grammars.
> > That is an oversight, and I'll fix it.
> >
> Thanks.

Glad I could clear this up. Thanks for raising the issue.

Eric Niebler
Boost Consulting

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