Hi,
maybe this question was already put it here, but can someone explain how
exactly work internal a function wrapped with lazy-seq keyword. For example
in the below code sample:
(
defn test-fc
"sum of all collection elements using recursion and laziness"
[coll]
(letfn [(sum-fc [sum coll]
(if (empty? coll)
(cons sum nil)
(lazy-seq(sum-fc (+ sum (first coll)) (rest coll))))
)
]
(sum-fc 0 coll)
)
)
if I test the function: (test-fc (range 5)) I got the right result, if I
continue to test with bigger number I don't got StackoverflowException, but
if run (test-fc (range 210432423543654675765876879)) I didn't get
StackoverflowEx but the application didn't return any result, because take
to much time to compute this ?
How exactly work this internally and how is removed recursion( inside call
sum-fc with new parameters) from the flow in this case ? From what I saw in
java code of LazySeq class and clojure source code, it's made a list with
LazySeq object and in my opinion in that list the LazySeq object contain
enough information to compute the requested item and in this way is removed
recursion and implicit StackoverflowException issue, if I understand
something wrong please explain to me.
I test this function from IntellijIDEA + LaClojure plugin.
Thanks
Sorin.
--
You received this message because you are subscribed to the Google
Groups "Clojure" group.
To post to this group, send email to clojure@googlegroups.com
Note that posts from new members are moderated - please be patient with your
first post.
To unsubscribe from this group, send email to
clojure+unsubscr...@googlegroups.com
For more options, visit this group at
http://groups.google.com/group/clojure?hl=en
---
You received this message because you are subscribed to the Google Groups
"Clojure" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to clojure+unsubscr...@googlegroups.com.
For more options, visit https://groups.google.com/d/optout.
