On Sun, Dec 5, 2010 at 5:16 PM, Robert McIntyre <r...@mit.edu> wrote:
> Your function never actually ends because even the empty list
> evaluates to true :)
>
> rlm.dna-melting> (if (rest '()) true false)
> true
>
> rlm.dna-melting> (if (next '()) true false)
> false
>
> so, changing your list length function to use next will work
>
> rlm.dna-melting> (defn list-length [col] (if col (+ 1 (list-length
> (next col))) 0))
> #'rlm.dna-melting/list-length
> rlm.dna-melting> (list-length (range 5))
> 5
>
> hope that helps,
> --Robert McIntyre
Yeah, but
user=> (list-length ())
1
:)
And the original code will still bomb on long enough lists. How about:
(defn list-length [coll]
(let [ll (fn [n s]
(if s
(recur (inc n) (next s))
n))]
(ll 0 (seq coll))))
user=> (list-length ())
0
user=> (list-length (range 5))
5
user=> (list-length (range 10000000))
10000000
The inner function takes a second argument that is the count-so-far.
This is so that the recursion can be turned into a tail call. As you
can see, no stack overflow even on a very long input.
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