Recursive lazy seq in Clojure

Question

I can't understand why this lazy-seq causes a stackoverflow, and why not when you pass the sequence to dorun:

(defn very-lazy [s]
    (lazy-seq
   (if (seq s)
     [(first s) (very-lazy (rest s))]
     [])))

(dorun (very-lazy (range 200000000)))
>nil

(take 2 (very-lazy (range 20000000))
>...(1577 (java.lang.StackOverflowError

If it's lazy then take 2 should cause the lazy seq to iterate only two times, why doesn't happen and why dorun works?

mobyte · Accepted Answer · 2012-12-22 14:31:41Z

3

In your example function returns lazyseq (0 (1 (2 (3 (...))))). That's why dorun runs without stackoverflow (there is sequence of two elements 0 and (1 (2 (...))) that dorun doesn't evaluate) and second fails (it returns infinite nested sequences that repl tries evaluate to print out).

I guess you're looking for this solution

(defn very-lazy [s]
  (lazy-seq
   (if (seq s)
     (cons (first s) (very-lazy (rest s)))
     [])))

(take 10 (very-lazy (range 200000000)))
-> (0 1 2 3 4 5 6 7 8 9)

edited Dec 22, 2012 at 14:31

answered Dec 22, 2012 at 12:54

mobyte

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2 Comments

Alex Over a year ago

Evaluating lazy seqs in the REPL often causes problems trying to print the results that would not occur in normal usage, e.g. just typing (repeat 1) in the REPL is a bad idea.

mobyte Over a year ago

You can set *print-length* to limit output, e.g. for (set! *print-length* 10) (repeat 1) will be outputted as (1 1 1 1 1 1 1 1 1 1 ...) in repl.

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