Go 中使用通道的斐波那契数列

Question

我正在按照 tour.golang.org 上的示例进行操作。

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 退出频道时它会停止？不管是否传递 0 来退出，x 总是有一个值。那么 select 不应该总是属于 case 'c <- x' 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，因为这个通道是无缓冲的，发送将被阻塞，直到有人可以从它接收。

了解 Effective Go 上的频道：

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 个案例可以继续，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.

Go 中使用通道的斐波那契数列

Fibonacci in Go using channels

fibonacci

go