go - Go 中使用 channel 的斐波那契数列

Question

我正在关注 tour.golang.org 上的示例。

我基本上理解这个例子，我唯一的问题是为什么当我们传递 0 退出 channel 时它会停止？不管是否传递 0 来退出，x 总是有一个值。所以 select 不应该总是落在 case 'c <- x' 上吗？

func fibonacci(c chan int, quit chan int) {
    x, y := 0, 1
    for {
        select {
        case c <- x:
            x, y = y, x+y
        case <-quit:
            return
        }
    }
    close(c)
}

func main() {
    c := make(chan int)
    quit := make(chan int)
    go func() {
        for i := 0; i < 10; i++ {
            fmt.Println(<-c)
        }
        quit <- 0
    }()
    fibonacci(c, quit)
}

Answer 1

there is always a value for x. So shouldn't select always fall on case 'c <- x' ?

不，因为这个 channel 是无缓冲的，发送将阻塞直到有人可以从它接收。

在 Effective Go 上阅读有关 channel 的信息:

Receivers always block until there is data to receive. If the channel is unbuffered, the sender blocks until the receiver has received the value. If the channel has a buffer, the sender blocks only until the value has been copied to the buffer; if the buffer is full, this means waiting until some receiver has retrieved a value.

此外，如果 select 语句中的 2 个 case 可以继续，one is picked pseudo-randomly .

If one or more of the communications can proceed, a single one that can proceed is chosen via a uniform pseudo-random selection. Otherwise, if there is a default case, that case is chosen. If there is no default case, the "select" statement blocks until at least one of the communications can proceed.