Hi,
With the new version without laziness, what exact test do you pass ?

Is it still (do (time (doall (to-list tree))) :done) ?

Because the doall adds an unfair re-traversal of the list, I think

2012/7/23 Alexander Semenov <[email protected]>

> I see. Just removed the lazyness and got ~800ms.
>
>
> (defn generate-tree [[h & t :as coll]]
>   (if (seq coll)
>     (let [lr (generate-tree t)] [h lr lr])
>     nil))
>
> (defn to-list [[v l r :as node]]
>   (if-not (nil? node)
>     (into
>      (to-list l)
>      (conj (to-list r) v))
>     []))
>
>
> On Monday, July 23, 2012 11:20:50 PM UTC+3, tbc++ wrote:
>>
>> > Thanks. But I don't do any number crunching here - just a huge
>> structure
>> > creation in memory. I can't get why removing the 'lazy-seq' wrapper
>> from the
>> > second 'concat' argument make things 10x times slower.
>>
>> Lazy-seqs require the allocation of a LazySeq object. Due to the lazy
>> nature of this structure, we have to protect against multiple
>> evaluation. This means that certain methods have to be synchronized:
>> (https://github.com/clojure/**clojure/blob/master/src/jvm/**
>> clojure/lang/LazySeq.java<https://github.com/clojure/clojure/blob/master/src/jvm/clojure/lang/LazySeq.java>).
>>
>> All that makes it quite a bit slower than a simple Iterator.
>>
>>
>> Timothy
>>
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