在例行程序开始后立即调用等待组完成？

Question

我在《Go Tools and Techniques for Developers 的 Concurrency in Go Tools and Techniques for Developers》中看到这段代码，其中提到了广播的用法，上下文是使用广播来唤醒三个 goroutings。

package main

import (
    "fmt"
    "sync"
    "time"
)

func main() {

    type Button struct {
        Clicked *sync.Cond
    }
    button := Button{Clicked: sync.NewCond(&sync.Mutex{})}

    subscribe := func(c *sync.Cond, fn func()) {
        var goroutineRunning sync.WaitGroup
        goroutineRunning.Add(1)
        go func() {
            goroutineRunning.Done() // <---- why here?
            //fmt.Println("wg already done")
            c.L.Lock()
            defer c.L.Unlock()
            c.Wait()
            fn()
            //goroutineRunning.Done(), if put here will result in deadlock, why?
            
        }()
        goroutineRunning.Wait()
    }

    var clickRegistered sync.WaitGroup
    clickRegistered.Add(3)

    subscribe(button.Clicked, func() {
        fmt.Println("Maximizing window.")
        clickRegistered.Done()
    })
    subscribe(button.Clicked, func() {
        fmt.Println("Displaying annoying dialog box!")
        clickRegistered.Done()
    })
    subscribe(button.Clicked, func() {
        fmt.Println("Mouse clicked.")
        clickRegistered.Done()
    })

    time.Sleep(time.Second * 3)
    button.Clicked.Broadcast()
    clickRegistered.Wait()

}

我正在尝试了解订阅部分

subscribe := func(c *sync.Cond, fn func()) {
        var goroutineRunning sync.WaitGroup
        goroutineRunning.Add(1)
        go func() {
            goroutineRunning.Done()
            //fmt.Println("wg already done")
            c.L.Lock()
            defer c.L.Unlock()
            c.Wait()
            fn()
            //goroutineRunning.Done()
            //fmt.Println("fn executed")
        }()
        goroutineRunning.Wait()
    }

作者说：

Here we define a convenience function that will allow us to register functions to handle signals from a condition. Each handler is run on its own goroutine, and subscribe will not exit until that goroutine is confirmed to be running.

我的理解是应该defer goroutingRunning.Done()在gorouting里面，这样后面的代码（包括等待Cond和fn()的调用）才有机会到运行), 但在这种情况下 goroutingRunning.Done() 似乎必须在 gorouting 的开头，否则会导致死锁错误，但为什么？

Answer 1

这是一种确保subscribe returns时goroutine已经启动运行ning的机制。当 goroutine 启动时，它会调用 Done 以向等待的调用者发出 goroutine 正在运行ning 的信号。如果没有这个机制，有可能在订阅的时候returns goroutine还没有被调度

延迟的Done将不起作用，因为它只会运行一次goroutine returns，并且只有在条件变量发出信号后才会发生。

该方案不保证新的协程锁定互斥体。这个模式是否真的有必要值得商榷。

在例行程序开始后立即调用等待组完成？

calling wait group done right after go routine starts?

synchronization

deadlock

go

waitgroup