Can anyone tell me why the following is true:
(setq a 1)
(setq b 2)
(setq c 3)
(setq d 1)
(case d
(a (do-this))
(b (do-that))
(c (do-something))
(true)
)
=> nil
I understand that the comparative values in my case are unevaluated. When switched out for an (eval) the case still doesn't work.
(setq a 1)
(setq b 2)
(setq c 3)
(setq d 1)
(case d
((eval a) (do-this))
((eval b) (do-that))
((eval c) (do-something))
(true)
)
=> nil
What don't I understand?
As this example shows, the expression in the exp-N position in each (exp-N body-N) pair is completely unevaluated.
> (set 'a 5)
5
> (set 'test '(eval a))
(eval a)
> (eval test)
5
> (case test (5 'zero) ((eval a) 'one) (pete 'two) (true 'magoo))
one
That is, no attempt is made to evaluate the expression (eval a) in ((eval a) 'one).
yes, you have to use constants:
(case d
(1 (do-this))
(2 (do-that))
(3 (do-something))
(true)
)
Lutz
Is there a work around, or can we impliment a change? Does it work this way for a reason?
Quote from: "statik"
Dmitry has written "ecase":
(eval (append (list 'case _v)
(map (fn (_i) (set-nth 0 _i (eval (_i 0))))
(args)))))
(set 'a 1)
(set 'b 2)
(ecase b
((eval a) (println "a is 1"))
((eval b) (println "b is 2"))
(true)
)
//http://en.feautec.pp.ru/SiteNews/funlib_lsp
'case' is a very fast switcher for the reason that is does not evaluate. For many years actually it was working
Lutz
Use 'if' if you want to evaluate...
Fanda