Hello,

(take 1 fib-seq) => (1)


Which can also be seen as[*] (map + (0) (1))

(map + '(0) '(1)) => (1)


Makes sense?

(take 2 fib-seq) => (1 1)


Here the recursive definition (note that it's not a function, fib-seq
simply a Var that holds a LazySeq object) comes into play. We already know
the first element in the sequence, so:

(map + '(0 0) '(1 1)) => (1 1)


The second 1 there is the first element of the fib-seq.


*: I'm using vague language intentionally here. Anyone who knows the right
term please correct me.


On Sun, Mar 9, 2014 at 12:54 PM, Asfand Yar Qazi <ayq...@gmail.com> 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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