Hi Juan, You cannot terminate a function execution from its caller if you did not pass it an argument (most likely a channel) which it can use to communicate with its caller, and terminate itself when the timeout is reached. The simplest example I can think of is the following: https://play.golang.org/p/KVQ7uhiWr7H, though I'm not quite sure this is your question, since it looks like you've already experimented with this construct.
You can trick with signals and other constructs but channels are designed for that kind of need, so I suppose this is the best way to go. If you want something that wraps this behaviour because you need the timeout to follow the logic of your program, maybe package https://golang.org/pkg/context/ can help you. Hope this was useful! Le mar. 25 févr. 2020 à 00:45, Juan Mamani <juanmamanichar...@gmail.com> a écrit : > Sorry again. Well, I followed your advice. Here is the link > https://play.golang.org/p/7DigEVsKbdx > How can I cancel goToSleep() when timeout is reached? > > > El lunes, 24 de febrero de 2020, 14:16:32 (UTC-3), Jake Montgomery > escribió: >> >> Your code will still not compile. In this group, it is often helpful to >> include a link to your code in the playground (https://play.golang.org/) >> using the "Share" button. That is in addition to, or instead of, posting >> your code in the message. This will allow others to easily run your code, >> but also will allow you to make sure it compiles. >> >> On Monday, February 24, 2020 at 10:01:01 AM UTC-5, Juan Mamani wrote: >>> >>> Sorry}, here is the right code: ( just was the effect to b working til >>> At 5:20am sleepy and lost) >>> >>> Expected behaviour is to end anonyimous func when 3s timeout is >>> reached. >>> >>> //------------------------------------------------ >>> package main >>> >>> import( >>> //"fmt" >>> "time" >>> "math/rand" >>> "log" >>> ) >>> >>> func main(){ >>> >>> for{ >>> log.Println("-----------------Start") >>> Task() >>> log.Println("-----------------End") >>> >>> } >>> >>> } >>> >>> >>> // Trying to adapt from Concurrency in Go by Katherine Cox-Buday >>> func Task(){ >>> >>> doWork := func( done <-chan interface{}, strings <-chan string, ) >>> <-chan interface{} { //1 >>> terminated := make(chan interface{}) >>> go func() { >>> defer log.Println("doWork exited.") >>> defer close(terminated) >>> for { >>> goToSleep() >>> select { >>> //case s := <-strings: >>> // case s := ran(): >>> // Do something interesting >>> //fmt.Println(s) >>> case <-done: //2 >>> return >>> } >>> } >>> }() >>> return terminated >>> } >>> >>> done := make(chan interface{}) >>> terminated := doWork(done, nil) >>> >>> go func() { //3 >>> // Cancel the operation after 3 second. >>> time.Sleep(3 * time.Second) >>> log.Println("Canceling doWork goroutine reach 3s...") >>> close(done) >>> }() >>> >>> <-terminated //4 >>> log.Println("Done.") >>> } >>> >>> done := make(chan interface{}) >>> terminated := doWork(done, nil) >>> >>> go func() { //3 >>> // Cancel the operation after 3 second. >>> time.Sleep(3 * time.Second) >>> log.Println("Canceling doWork goroutine reach 3s...") >>> close(done) >>> }() >>> >>> <-terminated //4 >>> log.Println("Done.") >>> } >>> >>> >>> >>> func goToSleep(){ >>> >>> >>> >>> >>> rand.Seed(time.Now().UnixNano()) >>> n := rand.Intn(12) // n will be between 0 and 10 >>> log.Println("Sleeping ",n,"seconds..") >>> time.Sleep(time.Duration(n)*time.Second) >>> log.Println("Done sleeping!") >>> } >>> >>> //------------------------------------------------ >>> >>> >>> >>> El lunes, 24 de febrero de 2020, 6:09:09 (UTC-3), Lutz Horn escribió: >>>> >>>> > I've tried to adapt from from "Concurrency in Go by Katherine >>>> Cox-Buday" to >>>> > understand how to apply timeout. But I don't get it. >>>> >>>> What are you trying to do? What is the expected behaviour, what happens >>>> instead? >>>> >>>> BTW, this code does not compile, there are same unbalanced curly >>>> brackets. >>>> >>>> Lutz >>>> >>> -- > You received this message because you are subscribed to the Google Groups > "golang-nuts" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to golang-nuts+unsubscr...@googlegroups.com. > To view this discussion on the web visit > https://groups.google.com/d/msgid/golang-nuts/b0828575-fddf-4024-bec0-8faf5b9858a4%40googlegroups.com > <https://groups.google.com/d/msgid/golang-nuts/b0828575-fddf-4024-bec0-8faf5b9858a4%40googlegroups.com?utm_medium=email&utm_source=footer> > . > -- You received this message because you are subscribed to the Google Groups "golang-nuts" group. To unsubscribe from this group and stop receiving emails from it, send an email to golang-nuts+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/golang-nuts/CANgi337rdhAK04LUKBnwR%3Dz1JnzyG9G_jmYMj1aqYJq%2BJzBFyw%40mail.gmail.com.
