lazy-seq only delays its own evaluation, it is not "recursive" :
(lazy-seq [0 1 2 3]) will evaluate the whole vector as soon as it is
forced. This means that it should wrap the tail of the lazy sequence
you are building.

Consequently, I find that the easiest way to use lazy-seq is as a
second argument to cons:

(def fib (cons 0 (cons 1 (lazy-seq (map + fib (rest fib))))))

but that might come from my playing with The Little Schemer.

On 1 May 2013 19:51, Stephen Compall <stephen.comp...@gmail.com> wrote:
> On Apr 30, 2013 5:32 AM, "Liao Pengyu" <arise...@gmail.com> wrote:
>>     (def fib
>>       (lazy-seq
>>         (concat [0 1] (map + fib (rest fib)))))
>>     (take 10 fib) ;; Bomb
>
> The expression (rest fib) forces fib, which is the lazy seq you are already
> trying to force when you eval (rest fib).
>
>>     (def fib
>>       (concat [0 1] (lazy-seq (map + fib (rest fib))))) ;; Works well
>
> Whereas here, fib has already been forced when you call (rest fib),
>
>>     (def fib
>>       (lazy-cat [0 1] (map + fib (rest fib)))) ;; Works well
>
> And here forcing fib doesn't eval (rest fib).
>
>> All of the above program using lazy-seq
>
> As you have seen, lazy-seq is not a silver bullet.
>
> --
> Stephen Compall
> If anyone in the MSA is online, you should watch this flythrough.
>
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