Thanks a lot for your explanation, the thing that confuses me was the same
function used with "repeatedly" only prints once:

(first (repeatedly 4 #(do (println "executed!") (inc 1))))

I guess because Clojure does not read-ahead in this scenario. Am I right?

On Sat, Oct 18, 2014 at 1:41 PM, Jonathan Winandy <
jonathan.wina...@gmail.com> wrote:

> You can try in you repl with
>
> (take 1 (map (fn [_]  (println "executed")) (vec (range 100))))
> (take 32 (map (fn [_]  (println "executed")) (vec (range 100))))
> (take 33 (map (fn [_]  (println "executed")) (vec (range 100))))
>
>
> you will observe the 32 sized chunks.
>
>
>
> On Sat, Oct 18, 2014 at 7:36 PM, James Reeves <ja...@booleanknot.com>
> wrote:
>
>> Some lazy lists are chunked for efficiency, which means Clojure will
>> read-ahead and evaluate a number of elements in advance. Often the outputs
>> from the various list handling functions are chunked (e.g. map, range,
>> etc.), while creating a seq explicitly with lazy-seq will not be chunked.
>>
>> - James
>>
>> On 18 October 2014 18:28, shahrdad shadab <clojure.langu...@gmail.com>
>> wrote:
>>
>>> Greeting everyone,
>>>
>>>  It might be stupid question but I expect
>>>
>>>  (first (map (fn [_]  (println "executed")) [1 2 3 4]))
>>>
>>> prints only once (realizing only first element in lazy seq returned by
>>> map) but it prints four times.
>>>  Can some one shed a light why?
>>>
>>> Thanks in advance
>>> Best regards
>>> Shahrdda
>>>
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