(case) evaluation

[statik]

Can anyone tell me why the following is true:

Code: Select all

(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.

Code: Select all

(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?

[Sammo]

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).

[Lutz]

yes, you have to use constants:

Code: Select all

(case d
   (1 (do-this))
   (2 (do-that))
   (3 (do-something))
   (true)
) 

Lutz

[statik]

Is there a work around, or can we impliment a change? Does it work this way for a reason?

[cormullion]

Is there a work around, or can we impliment a change? Does it work this way for a reason?

Dmitry has written "ecase":

Code: Select all

(define-macro (ecase _v)
  (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

[Lutz]

'case' is a very fast switcher for the reason that is does not evaluate. For many years actually it was working with evaluation, but then got changed to be more compatible with other LISPs.

Lutz

[Fanda]

Use 'if' if you want to evaluate...

Fanda