Go 最简单的竞走条件示例？

我需要一个简单的围棋代码样本，它肯定会运行到比赛条件下的程序

有什么想法吗？

原始问题：

我需要一个简单的围棋代码样本，它肯定会运行该程序 进入比赛状态

---

比如说,

racer.go

：

package main

import (
    "time"
)

var count int

func race() {
    count++
}

func main() {
    go race()
    go race()
    time.Sleep(1 * time.Second)
}

输出：

$ go run -race racer.go
==================
WARNING: DATA RACE
Read at 0x00000052ccf8 by goroutine 6:
  main.race()
      /home/peter/gopath/src/racer.go:10 +0x3a

Previous write at 0x00000052ccf8 by goroutine 5:
  main.race()
      /home/peter/gopath/src/racer.go:10 +0x56

Goroutine 6 (running) created at:
  main.main()
      /home/peter/gopath/src/racer.go:15 +0x5a

Goroutine 5 (finished) created at:
  main.main()
      /home/peter/gopath/src/racer.go:14 +0x42
==================
Found 1 data race(s)
exit status 66
$ 

---

运行代码时使用：

go Run-race.go

有两个goroutine，它们在相同的值上运行。这就是竞赛条件的定义。为了确保它确实遇到了数据竞争（你不能真正保证它，但已经足够接近），让他们在无限循环中对值进行操作。请不要在发布答案后将你的问题编辑成其他内容。相反，问一个新问题。我已将您的编辑回滚到已答复的版本。谢谢，这正是我想要的。

package main

import (
    "fmt"
)

func main() {
    i := 0
    // Run forever to make it to show race condition
    for {
        var x, y int

        // Set x to 60
        go func(v *int) {
            *v = 60
        }(&x)

        // Set y to 3
        go func(v *int) {
            *v = 3
        }(&y)

        /*
          A race condition is when multiple threads are trying to access and manipulate the same variable.
          the code below is all accessing and changing the value.
          Divide x (60) by y (3) and assign to z (42)...
          Except if y is not assigned 3 before x is assigned 60,
          y's initialized value of 0 is used,
          Due to the uncertainty of the Goroutine scheduling mechanism, the results of the following program is unpredictable,
          which causes a divide by zero exception.
        */
        go func(v1 int, v2 int) {
            fmt.Println(v1 / v2)
        }(x, y)

        i += 1

        fmt.Printf("%d\n", i)
    }
}