Thanks for your explanation which has allowed me to make this
definition
(def fibs
(list* 0 1 (lazy-seq (map + fibs (rest fibs)))))
that uses rest and IMHO is clearer that the first one using drop.
Juan Manuel
On 19 nov, 12:04, Christophe Grand <christo...@cgrand.net> wrote:
> Hi,
>
>
>
>
>
> On Fri, Nov 19, 2010 at 11:44 AM, babui <jmgim...@gmail.com> wrote:
> > I've been playing with lazy sequences defined by autoreferential
> > definition. For instance:
>
> > (def ones (lazy-seq (cons 1 ones)))
>
> > which is equivalent to (def ones (repeat 1)).
>
> > My problem arises when defining the sequence of fibonacci numbers.
> > With this definition:
>
> > (def fibs
> > (lazy-seq (list* 0 1 (map + fibs (drop 1 fibs)))))
>
> > all works, for instance:
>
> > user=> (take 10 fibs)
> > (0 1 1 2 3 5 8 13 21 34)
>
> > But if I change (drop 1 fibs) with (rest fibs), that is:
>
> > (def fibs
> > (lazy-seq (list* 0 1 (map + fibs (rest fibs)))))
>
> > and try to get the first 10, I get:
>
> > user=> (take 10 fibs)
> > java.lang.StackOverflowError
>
> > My problem is that I don't understand why.
>
> (drop 1 fibs) is fully lazy, it doesn't look at fibs to return a new seq
> (rest fibs) needs to evaluate the 1st item to return
> (next fibs) needs to evaluate the 1sy item and the rest (so most of the time
> it means evaluating the 2nd item too)
>
> (def fibs
> (lazy-seq (list* 0 1 (map + fibs (rest fibs)))))
>
> when you try to get the first item of the seq, it forces eval of (list* 0 1
> (map + fibs (rest fibs)))
> list* being a fn, 0, 1 and (map + fibs (rest fibs)) are evaluated
> map being a fn, +, fibs and (rest fibs) are evaluated
> and as as I mentioned above rest forces evaluation of the lazy-seq body (the
> list* call) which is exactly what we are actually doing, so we call
> ourselves until the stack blow.
>
> (rest 1 fibs) on the other hand delay the evaluation and saves us from the
> stack overflow.
>
> hth,
>
> Christophe
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