英文:
Why does this deadlock happen when I move a condition outside the select statement in Go
问题
我需要一个具有动态缓冲区的非阻塞通道，用于一个项目，所以我写了这段代码。
这是类型声明：
// receiver 是非阻塞通道的接收器
type receiver struct {
Chan <-chan string
list *[]string
mutex sync.RWMutex
}
// Clear 将缓冲区设置为0个元素
func (r *receiver) Clear() {
r.mutex.Lock()
*r.list = (*r.list)[:0]
r.mutex.Unlock()
// 丢弃剩余内容
if len(r.Chan) == 1 {
<-r.Chan
}
}
构造函数：
// NewNonBlockingChannel 返回非阻塞通道的接收器和发送器
func NewNonBlockingChannel() (*receiver, chan<- string) {
// 创建发送和接收通道以及缓冲区
send := make(chan string)
recv := make(chan string, 1)
list := make([]string, 0, 20)
r := &receiver{Chan: recv, list: &list}go func() {for {// 当接收器为空时，从缓冲区发送下一个元素if len(recv) == 0 && len(list) > 0 {r.mutex.Lock()recv <- list[len(list)-1]list = list[:len(list)-1]r.mutex.Unlock()}select {// 将传入的元素添加到缓冲区case s := <-send:r.mutex.Lock()list = append(list, s)r.mutex.Unlock()//default:}}}()return r, send
}
还有主函数中的测试代码：
func main() {
recv, sender := NewNonBlockingChannel()
// 向通道发送数据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) // 等待接收每个 "Hi"recv.Clear()for data := range recv.Chan {println(data)}
}
当我测试时，在接收来自发送器的 "case s := <-send:" 的 select 语句中发生了死锁，但是当我将发送下一个缓冲字符串的条件块移到 select 语句中时，一切都正常工作：
go func() {
for {select {// 将传入的元素添加到缓冲区case s := <-send:r.mutex.Lock()list = append(list, s)r.mutex.Unlock()default:// 当接收器为空时，从缓冲区发送下一个元素if len(recv) == 0 && len(list) > 0 {r.mutex.Lock()recv <- list[len(list)-1]list = list[:len(list)-1]r.mutex.Unlock()}}}
}()
我想知道为什么会出现这种情况。
英文:
I need a non blocking channel with dynamic buffer for a project so I wrote this code.
Here is the type declaration:
//receiver is the receiver of the non blocking channeltype receiver struct {Chan <-chan stringlist *[]stringmutex sync.RWMutex}//Clear sets the buffer to 0 elementsfunc (r *receiver) Clear() {r.mutex.Lock()*r.list = (*r.list)[:0]r.mutex.Unlock()//Discards residual contentif len(r.Chan) == 1 {<-r.Chan}}
Constructor:
//NewNonBlockingChannel returns the receiver & sender of a non blocking channelfunc NewNonBlockingChannel() (*receiver, chan<- string) {//Creates the send and receiver channels and the buffersend := 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 bufferif 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 buffercase s := <-send:r.mutex.Lock()list = append(list, s)r.mutex.Unlock()//default:}}}()return r, send}
And the toy test in the main:
func main() {recv, sender := NewNonBlockingChannel()//send data to the channelgo 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)}}
When I tested it and a deadlock happened in the select statement when I receive from the sender "case s := <-send:" but when I move the conditional block which sends the next buffered string into the select statement everything works perfectly:
go func() {for {select {//Adds the incoming elements to the buffercase s := <-send:r.mutex.Lock()list = append(list, s)r.mutex.Unlock()default://When the receiver is empty sends the next element from the bufferif len(recv) == 0 && len(list) > 0 {r.mutex.Lock()recv <- list[len(list)-1]list = list[:len(list)-1]r.mutex.Unlock()}}}}()
I'd like to know why.
答案1
得分: 0
假设有以下情景：
主程序发送了五个"bye"，其中两个被打印出来：
for data := range recv.Chan {println(data)}
剩下的三个被留在了list中（这是因为进入了三次for循环，条件len(recv) == 0 && len(list) > 0为假，所以只满足了case s := <-send，并且将s追加到了list中三次）。
而由NewNonBlockingChannel启动的goroutine正在等待语句s := <-send。没有更多的数据可以接收，所以它将永远等待下去。
当你将以下代码移动到select语句中时：
if len(recv) == 0 && len(list) > 0 {r.mutex.Lock()recv <- list[len(list)-1]list = list[:len(list)-1]r.mutex.Unlock()}
情况就不同了：
等待的goroutine将被唤醒，并且条件if len(recv) == 0 && len(list) > 0恰好为真。一切都进行得很顺利。
英文:
Imagine this scenario for example:
five "bye" has been send by the main, two of them has been print by
for data := range recv.Chan {println(data)}
three has been left in list (this is because three times enter for loop, the condition len(recv) == 0 && len(list) > 0 is false, so it just meet the case s := <-send, and do list = append(list, s) three times)
and the goroutine started by NewNonBlockingChannel is waiting on the statment s := <-send. no more staff to be receive, so it will wait forever
when you move the code
if len(recv) == 0 && len(list) > 0 {r.mutex.Lock()recv <- list[len(list)-1]list = list[:len(list)-1]r.mutex.Unlock()}
into the select, things are different:
the waiting gorountine will wake up, and the condition if len(recv) == 0 && len(list) > 0 is ture just in time. things goes well.
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