I have a function which processes a stream of values from a shared channel and looks for a value which satisfies a specific predicate:
(defn find-my-object [my-channel]
(go-loop []
(when-some [value (<! my-channel)]
(if (some-predicate? value)
value
(recur)
Now I want this function to return some 'failure' value after waiting for a timeout, which I have implemented like this:
(alts! [(find-my-object my-channel) (timeout 1000)])
The problem with this is that the go-loop above continues to execute after the timeout. I want find-my-object to drain values from my-channel while the search is ongoing, but I don't want to close my-channel on timeout, since it's used in some other places. What is the idiomatic way to implement this functionality?
Read from the timeout channel inside of find-my-object, not outside of it. You can hardcode that if you like, or make the function take a second channel argument, which is the "please stop reading things" channel. Then you can pass it a timeout channel to get the behavior you want now, or pass it some other kind of channel to control it in a different way.
Related
let's say I've got a channel out (chan). I need to take values that are put into the channel and add them. The number of values is undetermined (thus cannot use traditional loop with an end case with (<! out)) and comes from an external IO. I'm using a fixed timeout with alts! but that doesn't seem like the best way to approach the problem. So far, I've got the following (which I got from https://gist.github.com/schaueho/5726a96641693dce3e47)
(go-loop
[[v ch] (alts! [out (timeout 1000)])
acc 0]
(if-not v
(do (close! out)
(deliver p acc))
(do
(>! task-ch (as/progress-tick))
(recur (alts! [out (timeout 1000)]) (+ acc v)))))
The problem I've got is that a timeout of 1000 is sometimes not enough and causes the go-loop to exit prematurely (as it may take more than 1000ms for the IO operation to complete and put the val in the out channel). I do not think increasing the timeout value is such a good idea as it may cause me to wait longer than necessary.
What is the best way to guarantee all reads from the out channel and exit out correctly from the loop?
Update:
Why am I using timeout?
Because the number of values being put in the channel is not fixed; which means, I cannot create an exit case. W/o the exit case, the go-loop will park indefinely waiting ((<! out)) for values to be put in the channel out. If you have a solve without the timeout, that'd be really awesome.
How do i know I've read the last value?
I dont. That's the problem. That's why I'm using timeout and alts!! to exit the go-loop.
What do you want to do w/ the result?
Simple addition for now. However, that's not the important bit.
Update Final:
I figured out a way to get the number of values I'd be dealing with. So I modified my logic to make use of that. I'm still going to use the timeout and alts! to prevent any locking.
(go-loop
[[v _] (alts! [out (timeout 1000)])
i 0
acc 0]
(if (and v (not= n i))
(do
(>! task-ch (as/progress-tick))
(recur (alts! [out (timeout 1000)]) (inc i) (+ acc v)))
(do (close! out)
(deliver p* (if (= n i) acc nil)))))
I think your problem is a bit higher-up in your design, not a core-async specific one:
On one hand, you have an undetermined amount of values coming in a channel — there could be 0, there could be 10, there could be 1,000,000.
On the other hand, you want to read all of them, do some calculation and then return. This is impossible to do — unless there is some other signal that you can use to say "I think I'm done now".
If that signal is the timing of the values, then your approach of using alts! is the correct one, albeit I believe the code can be simplified a bit.
Update: Do you have access to the "upstream" IO? Can you put a sentinel value (e.g. something like ::closed) to the channel when the IO operation finishes?
The 'best' way is to wait for either a special batch ending message from out or for out to be closed by the sender to mark the end of the inputs.
Either way, the solution rests with the sender communicating something about the inputs.
Consider a core.async channel which is created like so:
(def c (chan))
And let's assume values are put and taken to this channel from different places (eg. in go-loops).
How would one flush all the items on the channel at a certain time?
For instance one could make the channel an atom and then have an event like this:
(def c (atom (chan))
(defn reset []
(close! #c)
(reset! c (chan)))
Is there another way to do so?
Read everything to a vector with into and don't use it.
(go (async/into [] c))
Let's define a little more clearly what you seem to want to do: you have code running in several go-loops, each of them putting data on the same channel. You want to be able to tell them all: "the channel you're putting values on is no good anymore; from now on, put your values on some other channel." If that's not what you want to do, then your original question doesn't make much sense, as there's no "flushing" to be done -- you either take the values being put on the channel, or you don't.
First, understand the reason your approach won't work, which the comments to your question touch on: if you deref an atom c, you get a channel, and that value is always the same channel. You have code in go-loops that have called >! and are currently parked, waiting for takers. When you close #c, those parked threads stay parked (anyone parked while taking from a channel (<!) will immediately get the value nil when the channel closes, but parked >!s will simply stay parked). You can reset! c all day long, but the parked threads are still parked on a previous value they got from derefing.
So, how do you do it? Here's one approach.
(require '[clojure.core.async :as a
:refer [>! <! >!! <!! alt! take! go-loop chan close! mult tap]])
(def rand-int-chan (chan))
(def control-chan (chan))
(def control-chan-mult (mult control-chan))
(defn create-worker
[put-chan control-chan worker-num]
(go-loop [put-chan put-chan]
(alt!
[[put-chan (rand-int 10)]]
([_ _] (println (str "Worker" worker-num " generated value."))
(recur put-chan))
control-chan
([new-chan] (recur new-chan)))))
(defn create-workers
[n c cc]
(dotimes [n n]
(let [tap-chan (chan)]
(a/tap cc tap-chan)
(create-worker c tap-chan n))))
(create-workers 5 rand-int-chan control-chan-mult)
So we are going to create 5 worker loops that will put their result on rand-int-chan, and we will give them a "control channel." I will let you explore mult and tap on your own, but in short, we are creating a single channel which we can put values on, and that value is then broadcast to all channels which tap it.
In our worker loop, we do one of two things: put a value onto the rand-int-chan that we use when we create it, or we will take a value off of this control channel. We can cleverly let the worker thread know that the channel to put its values on has changed by actually handing it the new channel, which it will then bind on the next time through the loop. So, to see it in action:
(<!! rand-int-chan)
=> 6
Worker2 generated value.
This will take random ints from the channel, and the worker thread will print that it has generated a value, to see that indeed multiple threads are participating here.
Now, let's say we want to change the channel to put the random integers on. No problem, we do:
(def new-rand-int-chan (chan))
(>!! control-chan new-rand-int-chan)
(close! rand-int-chan) ;; for good measure, may not be necessary
We create the channel, and then we put that channel onto our control-chan. When we do this, ever worker thread will have the second portion of its alt! executed, which simply loops back to the top of the go-loop, except this time, the put-chan will be bound to the new-rand-int-chan we just received. So now:
(<!! new-rand-int-chan)
=> 3
Worker1 generated value.
This gives us our integers, which is exactly what we want. Any attempt to <!! from the old channel will give nil, since we closed the channel:
(<!! rand-int-chan)
; nil
I have two unidirectional core.async channels :
channel out can only put!
channel in can only take!
And since this is ClojureScript the blocking operations are not available. I would like to make one bidirectional (in-out) channel out of those two (in and out).
(def in (async/chan))
(def out (async/chan))
(def in-out (io-chan in out)) ;; io or whatever the solution is
(async/put! in "test")
(async/take! ch (fn [x] (println x))) ;; should print "test"
(async/put! ch) ;; put into ch is equivalent to putting into `out`
I tried something like the following (not working) :
(defn io-chan [in-ch out-ch]
(let [io (chan)]
(go-loop []
(>! out-ch (<! io ))
(>! io (<! in-ch))
(recur))
io))
A schema might help :
out in-out
---------------> (unused)
<--------------- <---------------
in
----------------> ---------------->
<---------------- (unused)
Also, closing the bidirectional channel should close both underlying channels.
Is is possible ?
If I understand your use case right, I believe what you're trying to do is just a one-channel job.
On the other hand, if what you're trying to do is to present a channel-like interface for a composite of several channels (e.g some process takes data from in, processes it, and outputs the result to out), then you could always implement the right protocols (in the case of ClojureScript, cljs.core.async.impl.protocols/ReadPort and cljs.core.async.impl.protocols/WritePort).
I would personnaly not recommend it. Leaving aside the fact that you'd be relying on implementation details, I don't believe core.async channels are intended as encapsulation for processes, only as communication points between them. So in this use case, just pass the input channel to producers and the output channel to consumers.
Your example shows a flow basicly like this:
io ---> out-ch ---> worker ---> in-ch ---> io
^-------------------------------------------*
If we assume that worker reads from in-ch and writes to out-ch then perhaps these two channels are reversed in the example. if worker does the opposite then it's correct. in order to prevent loops it's important that you use non-buffered queues so you don't hear your own messages echoed back to yourself.
as a side note, there is no such thing as unidirectional and bi-directional channels. instead there are buffered and unbufferd channels. If we are talking over a buffered channel then when I have something to say to you, I park until you happen to be listening to the channel, then once you are ready to hear it I put my message into the channel and you receive it. Then to get a response I park until you are ready to send it, and once you are, you put it on the channel and I get it from the channel (all at once). This feels like a bi-directional channel though it's really just that unbuffered channels happen to coordinate this way.
If the channel if buffered then I might get my own message back from the channel, because I would finish putting it on the channel and then be ready to receive the response before you where even ready to receive the original message. If you need to use buffered channels like this then use two of them, one for each direction and they will "feel" like uni-directional channels.
I have followed here and am getting things called segments arriving at the aggregator. These segments all arrive and I can print them out as they arrive. But what I want to do is make an immutable data structure (a vector) out of them as they arrive. Or even wait for them all to arrive and then make the vector. I'll be able to know when the last one has arrived and sort them. I need to conj the arriving segment to the existing so-far-built-up vector. I'm used to creating vectors like this using returns from function calls, but I can't see how this facility is available to me in a thread or go block.
Another option is to use (clojure.core.async/into [] source-chan) . Like other "reducer-chans" in the async toolkit, it's also built on the assumption that you will close! the source channel when you're ready to receive the result.
Assumedly, since you are able to receive multiple items, your async code is already in a loop.
In order to build up a vector from the items you get, you should use a loop binding.
(def acc-chan
(>/go-loop [accumulator []]
(let [item (>/<! source-chan)]
(if (nil? item)
accumulator
(recur (conj accumulator item)))))
The go-loop call will immediately return a acc-chan, which will receive the loop's return value (the accumulator) when the loop exits. The accumulator is re-bound on each iteration of the loop, adding another item to the end. When source is closed, the accumulator is returned from the loop, and placed onto acc-chan where you can read it and use the value.
I'm looking at Clojure core.async for the first time, and was going through this excellent presentation by Rich Hickey: http://www.infoq.com/presentations/clojure-core-async
I had a question about the example he shows at the end of his presentation:
According to Rich, this example basically tries to get a web, video, and image result for a specific query. It tries two different sources in parallel for each of those results, and just pulls out the fastest result for each. And the entire operation can take no more than 80ms, so if we can't get e.g. an image result in 80ms, we'll just give up. The 'fastest' function creates and returns a new channel, and starts two go processes racing to retrieve a result and put it on the channel. Then we just take the first result off of the 'fastest' channel and slap it onto the c channel.
My question: what happens to these three temporary, unnamed 'fastest' channels after we take their first result? Presumably there is still a go process which is parked trying to put the second result onto the channel, but no one is listening so it never actually completes. And since the channel is never bound to anything, it doesn't seem like we have any way of doing anything with it ever again. Will the go process & channel "realize" that no one cares about their results any more and clean themselves up? Or did we essentially just "leak" three channels / go processes in this code?
There is no leak.
Parked gos are attached to channels on which they attempted to perform an operation and have no independent existence beyond that. If other code loses interest in the channels a certain go is parked on (NB. a go can simultaneously become a putter/taker on many channels if it parks on alt! / alts!), then eventually it'll be GC'd along with those channels.
The only caveat is that in order to be GC'd, gos actually have to park first. So any go that keeps doing stuff in a loop without ever parking (<! / >! / alt! / alts!) will in fact live forever. It's hard to write this sort of code by accident, though.
Caveats and exceptions aside, you can test garbage collection on the JVM at the REPL.
eg:
(require '[clojure.core.async :as async])
=> nil
(def c (async/chan))
=> #'user/c
(def d (async/go-loop []
(when-let [v (async/<! c)]
(println v)
(recur))))
=> #'user/d
(async/>!! c :hi)
=> true
:hi ; core.async go block is working
(import java.lang.ref.WeakReference)
=> java.lang.ref.WeakReference ; hold a reference without preventing garbage collection
(def e (WeakReference. c))
=> #'user/e
(def f (WeakReference. d))
=> #'user/f
(.get e)
=> #object[...]
(.get f)
=> #object[...]
(def c nil)
=> #'user/c
(def d nil)
=> #'user/d
(println "We need to clear *1, *2 and *3 in the REPL.")
We need to clear *1, *2 and *3 in the REPL.
=> nil
(println *1 *2 *3)
nil #'user/d #'user/c
=> nil
(System/gc)
=> nil
(.get e)
=> nil
(.get f)
=> nil
What just happened? I setup a go block and checked it was working. Then used a WeakReference to observe the communication channel (c) and the go block return channel (d). Then I removed all references to c and d (including *1, *2 and *3 created by my REPL), requested garbage collection, (and got lucky, the System.gc Javadoc does not make strong guarantees) and then observed that my weak references had been cleared.
In this case at least, once references to the channels involved had been removed, the channels were garbage collected (regardless of my failure to close them!)
Assumedly a channel produced by fastest only returns the result of the fastest query method and then closes.
If a second result was produced, your assumption could hold that the fastest processeses are leaked. Their results are never consumed. If they relied on all their results to be consumed to terminate, they wouldn't terminate.
Notice that this could also happen if the channel t is selected in the alt! clause.
The usualy way to fix this would be to close the channel c in the last go block with close!. Puts made to a closed channel will then be dropped then and the producers can terminate.
The problem could also be solved in the implementation of fastest. The process created in fastest could itself make the put via alts! and timeout and terminate if the produced values are not consumed within a certain amount of time.
I guess Rich did not address the problem in the slide in favor of a less lengthy example.