具有等待组和无缓冲通道的竞争条件

Question

在 post 中找到我最初问题的（正确）解决方案后，我想出了一个略有不同的解决方案（我认为读起来更好：

// 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)
}

它确实有效，但不幸的是运行它反对种族，它显示了 2 个种族条件：

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

你能帮我了解竞争条件在哪里吗？

Answer 1

您正在尝试从 fmt.Println(histogram) 中的 histogram 中读取，这与改变它的 goroutine 的写入不同步 histogram[w]++。您可以添加一个锁来同步写入和读取。

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)

或者干脆用一个频道等待。

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

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

fmt.Println(histogram)

具有等待组和无缓冲通道的竞争条件

Race condition with waitgroup and unbuffered channel

go

race-condition