Isn't that not only violating least astonishment, but potentially
introducing a terrible bug into the library? One could use "two different"
timeout channels in two different alts, and if the timeout channels are
aliased, then perhaps only one of the alts would actually get notified.
On Wed, Nov 20, 2013 at 5:05 AM, Michał Marczyk <michal.marc...@gmail.com>wrote:
> The reason = considers you timeout channels to be equal is that they
> are, in fact, the same object. In fact, they may end up being the same
> object even with different timeout values:
>
> (identical? (timeout 1) (timeout 2))
> ;= true
>
> Except I got a false just now with a fresh REPL and same timeout
> value... All subsequent invocations return true though, and I'm sure
> with a little digging the reason for the false would become clear.
>
> The reason for them being the same object is that timeout goes out of
> its way to avoid creating to many timeout channels and will reuse
> existing ones if their timeouts are due within a small amount of time
> (clojure.core.async.impl.timers/TIMEOUT_RESOLUTION_MS, currently 10
> ms) of the requested timeout.
>
> Cheers,
> Michał
>
> On 20 November 2013 10:08, Thomas G. Kristensen
> <thomas.g.kristen...@gmail.com> wrote:
> > Hi all,
> >
> > I ran into a core.async behaviour that confused me a bit the other day.
> In
> > some of our systems, we need to fire different timeouts, perform actions
> and
> > schedule a new timeout. The problem is, that if the timeouts are of the
> same
> > number of ms, we can't distinguish them, and therefore not keep track of
> and
> > remove them from a set (at least not easily).
> >
> > That sounds a bit fuzzy. Hopefully this spike will make it clearer what
> I'm
> > trying to say:
> >
> > (require '[clojure.core.async :refer [chan timeout alts!! alts!]])
> >
> > (= (chan) (chan))
> > ;; false
> >
> > (= (timeout 1) (timeout 2))
> > ;; false
> >
> > (= (timeout 1) (timeout 1))
> > ;; true
> >
> > (do (loop [ch->v (into {} (for [v [1 2 3]] [(timeout 1000) v]))]
> > (when-let [chs (keys ch->v)]
> > (let [[_ ch] (alts!! chs)]
> > (println (ch->v ch))
> > (recur (dissoc ch->v ch)))))
> > (println "done"))
> > ;; only fires "3", the last channel in the map
> >
> > The intended behaviour of the last loop is to print 1, 2 and 3 (not
> > necessarily in that order). However, the ch->v map will only contain one
> > key, as timeouts with the same duration are considered the same value. In
> > the real example, a new timeout with the same value should be scheduled
> > again, by being put in the map.
> >
> > So, my questions are:
> >
> > - Is this intended behaviour?
> > - Is there a different pattern for achieving the scheduling behaviour I'm
> > looking for?
> >
> > Thanks for your help,
> >
> > Thomas
> >
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