As for scheduling the printlns, this prints out 1, 2, 3 (in some order):
(let [c (chan)]
(doseq [v [1 2 3]]
(take! (async/timeout 1000) (fn [_] (put! c v))))
(go-loop [v (<! c)]
(println v)
(recur (<! c))))
Cheers,
Michał
On 20 November 2013 11:34, Michał Marczyk <michal.marc...@gmail.com> wrote:
> The behaviour of timeout channels is to close after the timeout
> elapses. This will be visible everywhere the channel is used.
>
> Cheers,
> Michał
>
>
> On 20 November 2013 11:22, Cedric Greevey <cgree...@gmail.com> wrote:
>> 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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