具有等待组和无缓冲通道的竞争条件

Question

在这篇文章Understanding golang channels: deadlock中得到了我最初问题的（正确的）解决方案之后，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 mainimport (    "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，但不幸的是在比赛中运行它，它显示了 2 个比赛条件：==================WARNING: DATA RACERead 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？