Convert list to variable arguments?

[Tim Johnson]

Let's suppose I have a function foo

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(define (foo)
    (doargs (i) (println i)))

and I can process it by something like this:

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(foo a b c x e f g)

Now, suppose I have a list that looks like this:

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(set 'lst '(a b c x e f g))

Is there a method to pass 'lst to 'foo so that it is consumend
one item at a time by 'doargs?
Thanks
tim

[Jeff]

(cons foo lst)

[Lutz]

like this:

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> (apply foo lst)
a
b
c
x
e
f
g
g
> 

[Tim Johnson]

(cons foo lst)

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> (cons foo lst)
((lambda () 
  (doargs (i) 
   (println i))) a b c x e f g)

;; That doesn't evaluate 'foo with 'lst...
;; Or did I miss something?
Thanks
tim

[Lutz]

we posted together ;-) the answer again:

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> (apply foo lst)
a
b
c
x
e
f
g
g
> 

[Tim Johnson]

we posted together ;-) the answer again:

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> (apply foo lst)
a
b
c
x
e
f
g
g
> 

Yes! Thank you sir!

[Lutz]

Jeff just forgot to wrap an 'eval' around his answer, but there is a difference between both approaches:

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(eval (cons foo lst))

This would give all nil's because a,b,c,d,e,f,g would be evaluated to their contents. It depends what you want, the contents of the symbols a to g or the symbols a to g themselves when using 'apply'.