我正在通过Go Concurrency Patterns: Pipelines and cancellation阅读/工作,但是在理解停止简短部分时遇到了麻烦。我们具有以下功能:
func sq(in <-chan int) <-chan int {
out := make(chan int)
go func() {
for n := range in {
out <- n * n
}
close(out)
}()
return out
}
func gen(nums ...int) <-chan int {
out := make(chan int)
go func() {
for _, n := range nums {
out <- n
}
close(out)
}()
return out
}
func merge(cs ...<-chan int) <-chan int {
var wg sync.WaitGroup
out := make(chan int, 1) // enough space for the unread inputs
// Start an output goroutine for each input channel in cs. output
// copies values from c to out until c is closed, then calls wg.Done.
output := func(c <-chan int) {
for n := range c {
out <- n
}
wg.Done()
}
wg.Add(len(cs))
for _, c := range cs {
go output(c)
}
// Start a goroutine to close out once all the output goroutines are
// done. This must start after the wg.Add call.
go func() {
wg.Wait()
close(out)
}()
return out
}
func main() {
in := gen(2, 3)
// Distribute the sq work across two goroutines that both read from in.
c1 := sq(in)
c2 := sq(in)
// Consume the first value from output.
out := merge(c1, c2)
fmt.Println(<-out) // 4 or 9
return
// Apparently if we had not set the merge out buffer size to 1
// then we would have a hanging go routine.
}
现在,如果您注意到
2
中的merge
行,它说我们制作了chan
大小为1的out buffer
,因为这对于未读的输入来说是足够的空间。但是,我几乎肯定我们应该分配一个chan
大小为2的buffer
。根据此代码示例:c := make(chan int, 2) // buffer size 2
c <- 1 // succeeds immediately
c <- 2 // succeeds immediately
c <- 3 // blocks until another goroutine does <-c and receives 1
因为此部分暗示
chan
大小为3的buffer
不会被阻塞。谁能澄清/帮助我理解? 最佳答案
该程序将两个值发送到out
通道,并从out
通道读取一个值。不会接收到其中一个值。
如果通道未缓冲(容量0),则发送的goroutine之一将阻塞,直到程序退出。这是泄漏。
如果创建的通道容量为1,则两个goroutine都可以发送到该通道并退出。发送到通道的第一个值由main
接收。第二个值保留在通道中。
如果main函数未从out
通道接收值,则需要一个容量为2的通道,以防止goroutine无限期阻塞。
关于go - Goroutines channel 和“止损”,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/34583722/