go - go-tour 解决方案中 binarytrees_quit.go 中的 quit channel 的目的是什么？

Question

我不太明白binarytrees_quit.go中quit channel 变量的用途。或者，我是否错过了这里的重点。我可以理解接收器可以发送值来退出以告诉 go 例程返回或退出。但我不认为这里是这种情况。是否只是为了确保 Walk 例程一直保留到 Same 完成执行？不会仅仅因为 channel 没有缓冲就例行公事。即使是这样，那也没有任何意义。请帮助我理解。

提前致谢!

Answer 1

你可以在“Go for gophers - GopherCon closing keynote - 25 April 2014 - Andrew Gerrand ”中看到详细的方法

Stopping early
Add a quit channel to the walker so we can stop it mid-stride.

func walk(t *tree.Tree, ch chan int, quit chan struct{}) {
    if t.Left != nil {
        walk(t.Left, ch, quit)
    }
    select {
    case ch <- t.Value:
    // vvvvvvvvvvvv
    case <-quit:
        return
    }
    // ^^^^^^^^^^^^
    if t.Right != nil {
        walk(t.Right, ch, quit)
    }
}

Create a quit channel and pass it to each walker.
By closing quit when the Same exits, any running walkers are terminated.

func Same(t1, t2 *tree.Tree) bool {
    // vvvvvvvvvvvv
    quit := make(chan struct{})
    defer close(quit)
    w1, w2 := Walk(t1, quit), Walk(t2, quit)
    // ^^^^^^^^^^^^
    for {
        v1, ok1 := <-w1
        v2, ok2 := <-w2
        if v1 != v2 || ok1 != ok2 {
            return false
        }
        if !ok1 {
            return true
        }
    }
}

安德鲁补充说:

Why not just kill the goroutines?

Goroutines are invisible to Go code. They can't be killed or waited on.
You have to build that yourself.

There's a reason:

As soon as Go code knows in which thread it runs you get thread-locality.
Thread-locality defeats the concurrency model.

• Channels are just values; they fit right into the type system.
• Goroutines are invisible to Go code; this gives you concurrency anywhere.

Less is more.