等待组和无缓冲通道的竞争条件

一尘不染

等待组和无缓冲通道的竞争条件

在本文中了解了我最初的问题的（正确）解决方案之后，了解了golang频道：死锁，我想出了一个略有不同的解决方案（在我看来，这更好看：

// Binary histogram counts the occurences of each word.
package main

import (
    "fmt"
    "strings"
    "sync"
)

var data = []string{
    "The yellow fish swims slowly in the water",
    "The brown dog barks loudly after a drink ...",
    "The dark bird bird of prey lands on a small ...",
}

func main() {
    histogram := make(map[string]int)
    words := make(chan string)
    var wg sync.WaitGroup
    for _, line := range data {
        wg.Add(1)
        go func(l string) {
            for _, w := range strings.Split(l, " ") {
                words <- w
            }
            wg.Done()
        }(line)
    }

    go func() {
        for w := range words {
            histogram[w]++
        }
    }()
    wg.Wait()
    close(words)

    fmt.Println(histogram)
}

它确实可以工作，但是不幸的是，它在比赛中运行，它显示了2个比赛条件：

==================
WARNING: DATA RACE
Read at 0x00c420082180 by main goroutine:
...
Previous write at 0x00c420082180 by goroutine 9:
...
Goroutine 9 (running) created at:
  main.main()

您能帮我了解比赛情况吗？

阅读 197

2020-07-02

共1个答案

一尘不染

您正在试图从读histogram中fmt.Println(histogram)未同步到够程变异它的写入histogram[w]++。您可以添加锁以同步写入和读取。

例如

var lock sync.Mutex

go func() {
    lock.Lock()
    defer lock.Unlock()
    for w := range words {
        histogram[w]++
    }
}()

//...
lock.Lock()
fmt.Println(histogram)

请注意，您也可以使用sync.RWMutex。

您可以做的另一件事是等待goroutine变异histogram完成。

var histWG sync.WaitGroup
histWG.Add(1)
go func() {
    for w := range words {
        histogram[w]++
    }
    histWG.Done()
}()

wg.Wait()
close(words)
histWG.Wait()

fmt.Println(histogram)

或者只是使用频道来等待。

done := make(chan bool)
go func() {
    for w := range words {
        histogram[w]++
    }
    done <- true
}()

wg.Wait()
close(words)
<-done

fmt.Println(histogram)

2020-07-02