Marko, you say "There is no doubling: *t* and *d* share the same underlying
lazy sequence and will refer to the same objects. The trouble is only that
you force the evaluation of *(count d)* while *(count t)* still waits to be
evaluated, so *t* must definitely stay bound to the head of the shared
sequence.".

But if there is no doubling, and single lazy sequence is in the memory in
both cases, how does then memory have problem with one case and not with
the other?
If both t and d refer to the same (realized) object in memory.

In both cases, to spit out t or d, the program must have it at one point in
its memory.

So what spends the EXTRA, critical, OOME memory in (count d) (count t) case?

Or does it get instantly garbaged the moment it gets realized in (count t)
(count d) case?

Anyway, thanks for the exhaustive discussion, Marko & Michal



On 18 April 2013 00:01, Michał Marczyk <michal.marc...@gmail.com> wrote:

> Note that the problem is not that t needs to hang around; it's that t
> holds a lazy sequence which hangs around in unrealized state. That
> lazy sequence internally holds a thunk -- a nullary function --
> capable of producing the actual sequence elements on request. It is
> this thunk that holds a reference to the underlying huge sequence.
> Once t is realized, the actual sequence gets cached and the thunk
> becomes eligible for GC (the field holding it is set to null). If it
> then needs to stay around for some other purpose, that is no problem:
>
> user=> (let [[t d] (split-with #(< % 12) (range 1e8))] [(count t)
> (count d) (count t)])
> [12 99999988 12]
>
> (Or I suppose you could return [(count d) (count t)], but (dorun t)
> before that.)
>
> Also, just to be explicit about this, calling (let [x
> (produce-huge-seq)] (count x)) is not a problem, because x gets
> cleared prior to control being handed off to count.
>
> I've also discussed the details of what's going on on SO, which is
> where I first noticed this question:
>
> http://stackoverflow.com/questions/15994316/clojure-head-retention
>
> Cheers,
> Michał
>
>
> On 17 April 2013 22:53, Marko Topolnik <marko.topol...@gmail.com> wrote:
> > On Monday, April 15, 2013 1:50:37 AM UTC+2, tyaakow wrote:
> >>
> >> Thank you for your response, Marko.
> >> I want to clarify one more thing:
> >>
> >> (let [[t d] (split-with #(< % 12) (range 1e8))]
> >>     [(count d) (count t)])
> >>
> >>
> >> does this mean that while (count d) is realizing (range 1e8) seq, it
> >> becomes (also) realized within t, therefore
> >> it doubles (range 1e8) in memory causing OOME while (count d) is still
> not
> >> finished?
> >
> >
> > There is no doubling: t and d share the same underlying lazy sequence and
> > will refer to the same objects. The trouble is only that you force the
> > evaluation of (count d) while (count t) still waits to be evaluated, so t
> > must definitely stay bound to the head of the shared sequence.
> >
> >>
> >> Also, you say "As count realizes one element after another, it doesn't
> on
> >> its own retain a reference to the past elements."
> >>
> >> Does this mean that, eg. in repl, when I do some (count xyz) and it
> >> realizes xyz, It will later need to be reevaluated (realized again) if I
> >> require xyz within repl (I presume that if I require xyz later within
> file,
> >> it wont be GC due to it and clojure will know it shouldnt be GC)
> >
> >
> > Be careful to observe that I say "doesn't on its own retain a reference
> to
> > the past elements". If you have xyz bound to the head of your sequence,
> it
> > will force the entire sequence to stay in memory for as long as xyz is
> > within scope (if it's a local) or indefinitely (if it's a global def'd
> var).
> > Generally, a lazy sequence never gets un-realized once it got
> realized---the
> > only option is for it to disappear entirely (turn into garbage).
> >
> > -marko
> >
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