英文:
goroutine leak with buffered channel in Go
问题
以下是《The Go Programming Language》中的代码：
func mirroredQuery() string {
responses := make(chan string, 3)
go func() { responses <- request("asia.gopl.io") }()
go func() { responses <- request("europe.gopl.io") }()
go func() { responses <- request("americas.gopl.io") }()
return <-responses // 返回最快的响应
}
func request(hostname string) (response string) { /* ... */ }
书中说：
如果我们使用了一个无缓冲的通道，那么两个较慢的 goroutine 将会在试图发送它们的响应到一个没有其他 goroutine 会接收的通道上而被阻塞。这种情况被称为 goroutine 泄漏，它是一个 bug。与垃圾变量不同，泄漏的 goroutine 不会自动回收，因此当不再需要时，确保 goroutine 自行终止非常重要。
问题是为什么这种情况会导致 goroutine 泄漏。在我看来，有缓冲的通道的容量是 3，3 个 goroutine 会发送它们的请求并立即退出，这不会导致泄漏。
英文:
Th following codes is in The Go Programming Language
func mirroredQuery() string {
responses := make(chan string, 3)
go func() { responses <- request("asia.gopl.io") }()
go func() { responses <- request("europe.gopl.io") }()
go func() { responses <- request("americas.gopl.io") }()
return <-responses // return the quickest response
}
func request(hostname string) (response string) { /* ... */ }
And book says
> Had we used an unbuffered channel, the two slower
> goroutines would have gotten stuck trying to send their responses
> on a channel from which no goroutine will ever receive . This situation, called a goroutine leak, would be a bug . Unlike garbage
> variables, leaked goroutines are not automatically collec ted, so
> it is important to make sure that goroutines terminate themselves
> when no longer needed.
<br/>
And the question is why this situation will cause a goroutine leak.In my idea, the buffered channel's cap is 3, and 3 goroutines will send their requests and exit immediately which will not cause the leak.
答案1
得分: 4
代码显示的部分不会导致泄漏。
正如段落所述：
> 如果我们使用一个非缓冲通道
意思是：如果我们使用了非缓冲通道...
所以只有当通道是非缓冲的时候，泄漏才会发生。
英文:
The code shown does not cause a leak.
As the paragraph states:
> Had we used an unbuffered channel
Meaning: if we had used an unbuffered channel...
So only if the channel was unbuffered the leak would occur.
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