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Re : set in Scheme
Sorry about the last message, I wrote the set macro down a little too
fast and it clearly won't work as intended. I meant to say something
(list 'eval (list 'quote (cadr e)))
(list 'quote (caddr e))))))
Then if you
(define a 's)
(define s 'any)
(define b 'r)
the effect of (set a b) should be to assign the value r to s.
So far so good. But what about this function?
(lambda (x y)
(set x y)))
If we (define s 'any) then the effect of (foo a b) is the same as above.
Judging by my last posting, this may not be perfect either, but at least it's
a lot closer. I'd be interested in seeing a shorter or better solution.
Thanks for your patience.
-- Brad Pierce