go - 当我在 Go 中将条件移到 select 语句之外时，为什么会发生此死锁

Question

我的项目需要一个带动态缓冲区的非阻塞 channel ，所以我编写了这段代码。

这是类型声明:

//receiver is the receiver of the non blocking channel
type receiver struct {
    Chan  <-chan string
    list  *[]string
    mutex sync.RWMutex
}

//Clear sets the buffer to 0 elements
func (r *receiver) Clear() {
    r.mutex.Lock()
    *r.list = (*r.list)[:0]
    r.mutex.Unlock()
    //Discards residual content
    if len(r.Chan) == 1 {
        <-r.Chan
    }
}

构造函数:

//NewNonBlockingChannel returns the receiver & sender of a non blocking channel
func NewNonBlockingChannel() (*receiver, chan<- string) {
    //Creates the send and receiver channels and the buffer
    send := make(chan string)
    recv := make(chan string, 1)
    list := make([]string, 0, 20)

    r := &receiver{Chan: recv, list: &list}

    go func() {

        for {
            //When the receiver is empty sends the next element from the buffer
            if len(recv) == 0 && len(list) > 0 {
                r.mutex.Lock()
                recv <- list[len(list)-1]
                list = list[:len(list)-1]
                r.mutex.Unlock()
            }

            select {
            //Adds the incoming elements to the buffer
            case s := <-send:
                r.mutex.Lock()
                list = append(list, s)
                r.mutex.Unlock()
                //default:

            }
        }
    }()

    return r, send
}

主要是玩具测试:

func main() {
    recv, sender := NewNonBlockingChannel()

    //send data to the channel
    go func() {
        for i := 0; i < 5; i++ {
            sender <- "Hi"
        }
        time.Sleep(time.Second)
        for i := 0; i < 5; i++ {
            sender <- "Bye"
        }
    }()
    time.Sleep(time.Millisecond * 70) //waits to receive every "Hi"
    recv.Clear()
    for data := range recv.Chan {
        println(data)
    }

}

当我测试它时，当我从发件人“case s := <-send:”收到时，选择语句中发生了死锁，但是当我移动发送下一个缓冲的条件 block 时将字符串放入 select 语句中一切正常:

go func() {

    for {
        select {
        //Adds the incoming elements to the buffer
        case s := <-send:
            r.mutex.Lock()
            list = append(list, s)
            r.mutex.Unlock()
        default:
            //When the receiver is empty sends the next element from the buffer
            if len(recv) == 0 && len(list) > 0 {
                r.mutex.Lock()
                recv <- list[len(list)-1]
                list = list[:len(list)-1]
                r.mutex.Unlock()
            }
        }
    }
}()

我想知道为什么。

Answer 1

想象一下这个场景:

五个“再见”已由主发送，其中两个已由打印

for data := range recv.Chan {
      println(data)
}

三个已留在list (这是因为三次进入for循环，条件len(recv) == 0 && len(list) > 0为假，所以刚好满足case s := <-send，做三次list = append(list, s))

goroutine 由 NewNonBlockingChannel 启动正在等待声明s := <-send .没有更多的工作人员可以接收，所以它将永远等待

移动代码时

if len(recv) == 0 && len(list) > 0 {
    r.mutex.Lock()
    recv <- list[len(list)-1]
    list = list[:len(list)-1]
    r.mutex.Unlock()
}

进入select , 事情是不同的:

等待的gorountine会被唤醒，条件if len(recv) == 0 && len(list) > 0来得正是时候。事情进展顺利。