与等待组和无缓冲通道的争用条件

Question

在这篇文章中为我的初始问题找到(正确的)解决方案后 了解 golang 渠道：僵局，我想出了一个稍微不同的解决方案(在我看来读起来更好：

// 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，它针对种族运行它，它显示了 2 个竞争条件：

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

你能帮我了解比赛条件在哪里吗？

Answer 1

您正在尝试从fmt.Println(histogram)中读取histogram，该与将其histogram[w]++突变的goroutine的写入同步。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。

你可以做的另一件事是等待goroutine突变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)

或者干脆使用通道等待。

done := make(chan bool)
go func() {
for w := range words {
histogram[w]++
}
done <- true
}()
wg.Wait()
close(words)
<-done
fmt.Println(histogram)