在单独的 go 例程中重置计时器

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问题内容

TimeOutTime在以下场景中，网络实体在执行特定任务之前总是等待几秒钟X。假设这个时间为TimerT。TimeOutTime如果实体收到一组外部消息，则在此等待几秒钟期间，它应该再次将其重置TimerT为TimeOutTime。如果没有收到外部消息，则预期行为如下：

1. 计时器已过期
2. 执行任务 X
3. 再次将计时器重置为TimeOutTime

（reset我的意思是，停止计时器并重新开始）

为了模拟场景，我在 Go 中编写了以下代码。

package main

import (
    "log"
    "math/rand"
    "sync"
    "time"
)

const TimeOutTime = 3
const MeanArrivalTime = 4

func main() {
    rand.Seed(time.Now().UTC().UnixNano())
    var wg sync.WaitGroup
    t := time.NewTimer(time.Second * time.Duration(TimeOutTime))
    wg.Add(1)
    // go routine for doing timeout event
    go func() {
        defer wg.Done()
        for {
            t1 := time.Now()
            
<p><code>-race</code>使用标志运行此程序后，它显示<code>DATA_RACE</code>：</p>
<pre class="brush:go;toolbar:false">==================
WARNING: DATA RACE
Write at 0x00c0000c2068 by goroutine 8:
  time.(*Timer).Reset()
      /usr/local/go/src/time/sleep.go:125 +0x98
  main.main.func1()
      /home/deka/Academic/go/src/main/test.go:29 +0x18f

Previous write at 0x00c0000c2068 by goroutine 9:
  time.(*Timer).Reset()
      /usr/local/go/src/time/sleep.go:125 +0x98
  main.main.func2()
      /home/deka/Academic/go/src/main/test.go:42 +0x80

Goroutine 8 (running) created at:
  main.main()
      /home/deka/Academic/go/src/main/test.go:20 +0x1d3

Goroutine 9 (running) created at:
  main.main()
      /home/deka/Academic/go/src/main/test.go:35 +0x1f5
==================

然后我使用 Mutex 将Reset()调用包装在 Mutex 中。

包主

import (
    "log"
    "math/rand"
    "sync"
    "time"
)

const TimeOutTime = 3
const MeanArrivalTime = 4

func main() {
    rand.Seed(time.Now().UTC().UnixNano())
    var wg sync.WaitGroup
    t := time.NewTimer(time.Second * time.Duration(TimeOutTime))
    wg.Add(1)
    var mu sync.Mutex
    // go routine for doing timeout event
    go func() {
        defer wg.Done()
        for {
            t1 := time.Now()
            
<p>根据以下观察，此代码似乎可以正常工作。</p>
<p>如果我更换线路</p>
<p><code>time.Sleep(time.Second * time.Duration(rand.Intn(MeanArrivalTime)))</code></p>
<p>在第二个 go 例程中，睡眠时间为 ，<code>4 seconds</code>并且<code>TimeOutTime</code>为<code>3 seconds</code>。</p>
<p>程序的输出是：</p>
<pre class="brush:go;toolbar:false">2020/02/29 20:10:11 Timeout after  3.000160828s
2020/02/29 20:10:15 Timeout after  4.000444017s
2020/02/29 20:10:19 Timeout after  4.000454657s
2020/02/29 20:10:23 Timeout after  4.000304877s

在上面的执行中，2ndgo 例程active timer在计时器花费初始一秒后重置。因此，从第二次打印开始的几秒钟timer后，它就会过期。4

现在，当我检查文档时，Reset()我发现以下内容：

// Copyright 2009 The Go Authors. All rights reserved.
// Use of this source code is governed by a BSD-style
// license that can be found in the LICENSE file.



// Reset changes the timer to expire after duration d.
// It returns true if the timer had been active, false if the timer had
// expired or been stopped.
//
// Reset should be invoked only on stopped or expired timers with drained channels.
// If a program has already received a value from t.C, the timer is known
// to have expired and the channel drained, so t.Reset can be used directly.
// If a program has not yet received a value from t.C, however,
// the timer must be stopped and鈥攊f Stop reports that the timer expired
// before being stopped鈥攖he channel explicitly drained:
//
//  if !t.Stop() {
//      
<p>我找到了这张图：（链接：[https](https://blogtitle.github.io/go-advanced-concurrency-
patterns-part-2-timers/) ://blogtitle.github.io/go-advanced-concurrency-
patterns-part-2-timers/ ）</p>
<p><img src="./page_107_21_files/owEnr.png" alt="时间图 GoLang"></p>
<p>考虑到图表，我似乎需要使用，</p>
<pre class="brush:go;toolbar:false">if !t.Stop() {
    
<p>在<code>2nd</code>go 例程中。在这种情况下，我还需要在两个 go 例程中进行适当的锁定，以避免在通道上无限等待。</p>
<p>我不明白<code>t.Stop() + draining of the channel (应该在什么情况下执行。在什么情况下需要？在我的示例中，我不使用通道读取值。我可以在不调用 Stop() 的情况下调用 Reset() 吗？</code></p>
<h2 class="daan">
    正确答案
</h2>
<p><code>time.After</code>我使用函数简化了代码：</p>
<pre class="brush:go;toolbar:false">package main

import (
    "log"
    "math/rand"
    "time"
)

const TimeOutTime = 3
const MeanArrivalTime = 4

func main() {
    const interval = time.Second * TimeOutTime
    // channel for incoming messages
    var incomeCh = make(chan struct{})

    go func() {
        for {
            // On each iteration new timer is created
            select {
            case 
<p>注意：</p>
<blockquote>
<p><code>After</code>等待持续时间过去，然后在返回的通道上发送当前时间。它相当于<code>NewTimer(d).C</code>。在定时器触发之前，垃圾收集器不会恢复底层定时器。如果效率是一个问题，请改用并在不再需要计时器时<code>NewTimer</code>
调用。<code>Timer.Stop</code></p>
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