"Z.A" writes:
> Meikel : To get the sequence printing logic i tried following code. Makes
> no sense to me.
>
>(def lazy2 (take 3 (iterate #(do
> (cond
> (= 0 %) (print "+")
> (= 1 %) (p
Thanks everybody.
Meikel : To get the sequence printing logic i tried following code. Makes
no sense to me.
(def lazy2 (take 3 (iterate #(do
(cond
(= 0 %) (print "+")
(= 1 %) (print "*")
A form like:
(def lazy2 (map #(str "+" %) (range)))
would work.
Le mercredi 6 juin 2012 11:24:06 UTC+2, Z.A a écrit :
>
> Hi
>
> user=> (def lazy1(take3 (iterate#(do (print "+")
> (inc %))0)))
> #'user/lazy1
> user=> lazy1
> (+0+1 2)
>
> Why am I not gettin
Hi,
Am Mittwoch, 6. Juni 2012 12:13:56 UTC+2 schrieb Jim foo.bar:
Your f is NOT free of side-effects...Nonetheless, I would expect (0 +1
> +2) instead of (+0 +1 2)!
> Can anyone shine some light on this?
>
>
When printing the sequence the tail is realized before the current item is
printed
This is due to the fact that the evaluation of a do form returns the last
expr...
Le mercredi 6 juin 2012 12:23:07 UTC+2, Jim foo.bar a écrit :
>
> It's obvious that you're looking for:
>
> (def lazy1 (interleave (repeat '+) (range 3)))
>
>
> but I still haven't figured out why you get: (+0 +
It's obvious that you're looking for:
(def lazy1 (interleave (repeat '+) (range 3)))
but I still haven't figured out why you get: (+0 +1 2) when using iterate!
Jim
On 06/06/12 11:13, Jim - FooBar(); wrote:
From the docs:
clojure.core/iterate
([f x])
Returns a lazy sequence of x, (f x
From the docs:
clojure.core/iterate
([f x])
Returns a lazy sequence of x, (f x), (f (f x)) etc. f must be free of
side-effects
Your f is NOT free of side-effects...Nonetheless, I would expect (0 +1
+2) instead of (+0 +1 2)!
Can anyone shine some light on this?
Jim
On 06/06/12 10:24, Z
Hi,
because the first element is taken verbatim and the function is not applied
to it. In this case it is the zero.
(iterate f x) => (x (f x) (f (f x)) ...)
So the pluses you see are for the 1 and the 2.
Kind regards
Meikel
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