Hi,

On Sunday, 9 March 2014 14:58:47 UTC, Atamert Ölçgen wrote:
>
> Hello,
>
> (take 1 fib-seq) => (1)
>
>
> Which can also be seen as[*] (map + (0) (1))
>
> (map + '(0) '(1)) => (1)
>
>
> Makes sense?
>
>
I'm afraid it still doesn't make sense; I still don't understand how (cons 
0 (cons 0 fib-seq)) evaluates to '(0) the first time; it should evaluate to 
'(0 0) since there are 2 cons statements.  That's the point I'm stuck on; 
any help would be gratefully received.





On Sun, Mar 9, 2014 at 12:54 PM, Asfand Yar Qazi <ayq...@gmail.com<javascript:>
> > wrote:
>
>> Hi,
>>
>> I'm trying to understand the following function (from 
>> http://en.wikibooks.org/wiki/Clojure_Programming/Examples/Lazy_Fibonacci#Self-Referential_Version
>> ):
>>
>> (def fib-seq
>>   (lazy-seq
>>     (map +
>>       (cons 0 (cons 0 fib-seq))
>>       (cons 1 fib-seq))))
>>
>> I'm trying to understand how this works.  In particular, I do not 
>> understand what the recursive call to fib-seq will return when the sequence 
>> is lazily evaluated.
>>
>> Here's my understanding so far:
>>
>> The first time fib-seq is invoked, it has no head, and the function is 
>> the tail.  So we go into the first collection, where we append 0 and 0 to 
>> fib-seq, which then '(0 0) .  This is then mapped with the second 
>> collection, which (because fib-seq has not returned anything yet) is '(1) . 
>>  Shouldn't map then raise an error because it is effectively being called 
>> as (map + (0 0) (1)) ?
>>
>> I would be very grateful for any insights.
>>
>> Thanks
>>
>>
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>
>
>
> -- 
> Kind Regards,
> Atamert Ölçgen
>
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> --+
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>
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>  

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