我试图了解频道在golang中的工作方式。我拥有的代码非常简单,但输出令人惊讶。
正如文档所述:从通道进行读写操作会阻塞当前的goroutine,因此我认为写入通道会阻塞该通道,直到主例程产生为止。
package main
func rtn(messages chan<- string) {
defer close(messages)
println("p1")
messages <- "ping1"
//for i := 0; i < 10000000; i++ { }
println("p2")
messages <- "ping2"
}
func main() {
messages := make(chan string)
go rtn(messages)
for msg := range messages {
println(msg)
}
}
我认为它将打印
p1
ping1
p2
ping2
但实际上可以打印
p1
p2
ping1
ping2
答案 0 :(得分:2)
您正在使用无缓冲通道,该通道用作主goroutine与第二goroutine之间的同步点。
在这种情况下,您仅知道当第二个goroutine在messages <- "ping1"
处时,主要的例程在for msg := range messages
行处。因此,不能保证主循环立即到达println(msg)
。也就是说,与此同时,第二个goroutine可能继续前进并到达了println("p2")
和messages <- "ping2"
行。
作为反例,我添加了一个通道以强制在打印之间完全同步。
package main
func rtn(messages chan<- string, syncChan chan struct{}) {
defer close(messages)
println("p1")
messages <- "ping1"
//Wait for main goroutine to print its message
<-syncChan
//for i := 0; i < 10000000; i++ { }
println("p2")
messages <- "ping2"
//Wait for main goroutine to print its message
<-syncChan
}
func main() {
messages := make(chan string)
syncChan := make(chan struct{})
go rtn(messages, syncChan)
for msg := range messages {
println(msg)
//Notify the second goroutine that is free to go
syncChan <- struct{}{}
}
}
输出您期望的输出:
p1
ping1
p2
ping2
这里是产生所需输出的另一个示例。在这种情况下,主goroutine被time.Sleep()
强制阻止。这将使第二个goroutine准备好在接收器准备好接收之前发送。因此,发送者实际上将阻止发送操作。
package main
import (
"time"
)
func rtn(messages chan<- string) {
defer close(messages)
println("p1")
messages <- "ping1"
//for i := 0; i < 10000000; i++ { }
println("p2")
messages <- "ping2"
}
func main() {
messages := make(chan string)
go rtn(messages)
//Put main goroutine to sleep. This will make the
//sender goroutine ready before the receiver.
//Therefore it will have to actually block!
time.Sleep(time.Millisecond * 500)
for msg := range messages {
println(msg)
}
}