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Mike Champion wrote:
> Actually, since S-expressions are isomorphic to the basic XML data model,
> one could argue that X# == LISP :-)
I've heard this claim asserted many, many times,
but I still don't believe it.
Can anyone demonstrate the isomorphism? That is,
come up with a pair of functions f :: S-Exp -> XML
and g :: XML -> S-Exp such that (f . g) = id_{XML}
and (g . f) = id_{S-Exp}?
Note that any such pair of functions must also satisfy
h(x) = f(g(h(x))) = h(f(g(x))) for all x :: XML and
h :: XML -> XML, including, for example,
h = XPath(ancestor::*/following-sibling::*).
--Joe English
jenglish@flightlab.com
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