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 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，但不幸的是在比赛中运行它，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()

你能帮我了解比赛条件在哪里吗？