如果已经运行，则不应执行 Ticker

Question

我有一些必须定期运行的功能。我为此使用了自动收报机。但是如果代码已经在运行，并且时间间隔再次过去，它不应该再次执行。package mainimport (    "fmt"    "time")func main() {    ticker := time.NewTicker(3*time.Second)    flag := 0    defer ticker.Stop()    for {        select {        case t := <-ticker.C:            flag = flag + 1            if (flag % 2 ==0 ) {                time.Sleep(time.Second*4)            }               fmt.Println("Current time: ", t)        }    }}https://play.golang.org/p/2xV2MYInn4I在操场上，自动收报机每 3 秒打印一次，但每次自动收报机的偶数转一圈，作业所花费的时间都比间隔时间长。我希望它不会运行并放弃那些滴答声。我该怎么做呢？

Answer 1

睡在同一个 goroutine 中只会延迟执行。同时，ticker 在一个单独的 goroutine 中运行。因此，即使您使用全局变量来维持执行状态 - 它也不会为您提供所需的睡眠结果。然而，在一个单独的 goroutine 中迁移整个“睡眠”会产生：

package main

import (

    "fmt"

    "time"

)

type Tick struct {

    ticker *time.Ticker

    executing bool

}

func somethingYouWantToDo(tick *Tick, flag *int, t time.Time) {

    if tick.executing {

        return

    }

    tick.executing = true

    *flag = *flag + 1

    if (*flag % 2 ==0 ) {

                time.Sleep(time.Second*4)

    }   

        fmt.Println("Current time: ", t)

    tick.executing = false

}

func main() {

    tick := &Tick{

        ticker: time.NewTicker(3*time.Second),

    }

    flag := 0

    defer tick.ticker.Stop()

    for {

        select {

        case t := <-tick.ticker.C:

            go somethingYouWantToDo(tick, &flag, t)

        }

    }

}

// output

// Current time:  2009-11-10 23:00:03 +0000 UTC m=+3.000000001

// Current time:  2009-11-10 23:00:06 +0000 UTC m=+6.000000001

// Current time:  2009-11-10 23:00:12 +0000 UTC m=+12.000000001

// Current time:  2009-11-10 23:00:15 +0000 UTC m=+15.000000001

// Current time:  2009-11-10 23:00:21 +0000 UTC m=+21.000000001

// Current time:  2009-11-10 23:00:24 +0000 UTC m=+24.000000001

// Current time:  2009-11-10 23:00:30 +0000 UTC m=+30.000000001

// Current time:  2009-11-10 23:00:33 +0000 UTC m=+33.000000001

// Current time:  2009-11-10 23:00:39 +0000 UTC m=+39.000000001

// Current time:  2009-11-10 23:00:42 +0000 UTC m=+42.000000001

// Current time:  2009-11-10 23:00:48 +0000 UTC m=+48.000000001

// Current time:  2009-11-10 23:00:51 +0000 UTC m=+51.000000001

// Current time:  2009-11-10 23:00:57 +0000 UTC m=+57.000000001

// Current time:  2009-11-10 23:01:00 +0000 UTC m=+60.000000001

// Current time:  2009-11-10 23:01:06 +0000 UTC m=+66.000000001

// Current time:  2009-11-10 23:01:09 +0000 UTC m=+69.000000001

Answer 2

代码通道是缓冲的，这就是为什么您可能会一个接一个地看到多个触发器。您可以通过简单地将代码的值传输到无缓冲通道来防止这种情况（另请注意，从代码接收的 time.Time 值不是当前时间，而是最后一次滴答的时间）：

package main

import (

    "fmt"

    "time"

)

func main() {

    c := make(chan time.Time) // unbuffered

    ticker := time.NewTicker(3 * time.Second)

    defer ticker.Stop()

    go func() {

        for t := range ticker.C {

            select {

            case c <- t:

            default:

            }

        }

    }()

    for flag := 0; flag < 8; flag++ {

        <-c

        if flag%2 == 0 {

            time.Sleep(time.Second * 4)

        }

        fmt.Println("Current time: ", time.Now())

    }

}

// Output:

// Current time:  2020-02-19 12:21:57.095433032 +0100 CET m=+3.000213350

// Current time:  2020-02-19 12:22:04.095585208 +0100 CET m=+10.000365520

// Current time:  2020-02-19 12:22:06.095363327 +0100 CET m=+12.000143680

// Current time:  2020-02-19 12:22:13.095605268 +0100 CET m=+19.000385598

// Current time:  2020-02-19 12:22:15.095371885 +0100 CET m=+21.000152174

// Current time:  2020-02-19 12:22:22.095537562 +0100 CET m=+28.000317857

// Current time:  2020-02-19 12:22:24.095431317 +0100 CET m=+30.000211625

// Current time:  2020-02-19 12:22:31.095524308 +0100 CET m=+37.000304595

在操场上试一试：https: //play.golang.org/p/jDe5uJiRVe2