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From: Larry Evans (cppljevans_at_[hidden])
Date: 2008-03-20 18:59:28

On 03/20/08 16:53, Eric Niebler wrote:
> Larry Evans wrote:
>> What about recording the arity *with* the tag instead of in
>> the expression associated with the tag? IOW, instead of:
>> expr<Tag,Args,Arity>
>> have:
>> expr<tag_arity<Tag,Arity>,Args>
>> Why? Well, it directly associates the arity with the Tag instead
>> of with the expression. After all, sin(X) is not expressed
>> as sin(x,1). [or in lisp terms, (sin x) is not expressed as
>> (sin x 1) ].
> What problem does that solve?

I don't think it solves any problem that can't
already be solved, but it makes the solution
clearer. It's clearer because the arity
is associated with the tag *not* the
expression. The arity of a tag does not depend
on the expression; so, it should not be in
the expression. Instead the arity of the tag
determines the expression validity.

OK, but you may say "the arity *still* is in
the expression:


"; however, it's "more closely" associated
with the tag than the expression, so it's
really more a matter of style, I guess.
OTOH, I would guess it would make specifying
morphisms between algebra's easier since
any morphism would map equal arity's to
equal arities; and if the arity were
directly associate with the tag, as in
tag_arity<Tag,Arity>, then the mapping
would be easier to understand and verify
(with maybe a boost concepts class
which assured the pairs in the mpl map
had equal arities on both sides).

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