在下面的代码中有两个包含工作的 channel A 和 B,在实际代码中它们是不同的结构,工作人员在退出前需要排空两个 channel 。工作人员需要来自两个 channel 的信息。这两个 select 语句有效,但非常笨拙。如果我添加 default:
使它们成为非阻塞的,那么代码将无法耗尽 channel 。有没有更好的方式来编写选择?
现在,如果 channel A 没有工作,那么 channel B 也不会得到服务。另一个需要解决的问题,但不是我主要关心的问题。
playground用于测试以下代码:
package main
import (
"fmt"
"time"
)
const (
fillCount = 10 // number of elements in each input channel
numWorkers = 3 // number of consumers.
)
func Wait() {
time.Sleep(2000 * time.Millisecond)
}
func fillChannel(work chan string, name string) {
for i := 0; i < fillCount; i++ {
work <- fmt.Sprintf("%s%d", name, i)
}
close(work) // we're finished
}
func doWork(id int, ch1 chan string, ch2 chan string, done chan bool) {
fmt.Println("Running worker", id)
defer fmt.Println("Ending worker", id)
for ch1Open, ch2Open := true, true; ch1Open && ch2Open; {
cnt1 := len(ch1)
cnt2 := len(ch2)
if ch1Open {
select {
case str, more := <-ch1:
if more {
fmt.Printf("%d: ch1(%d) %s\n", id, cnt1, str)
} else {
fmt.Printf("%d: ch1 closed\n", id)
ch1Open = false
}
}
}
if ch2Open {
select {
case str, more := <-ch2:
if more {
fmt.Printf("%d: ch2(%d) %s\n", id, cnt2, str)
} else {
fmt.Printf("%d: ch2 closed\n", id)
ch2Open = false
}
}
}
}
done <- true
}
func main() {
a := make(chan string, 2) // a small channel
b := make(chan string, 5) // a bigger channel
// generate work
go fillChannel(a, "A")
go fillChannel(b, "B")
// launch the consumers
done := make(chan bool)
for i := 0; i < numWorkers; i++ {
go doWork(i, a, b, done)
}
// wait for the goroutines to finish.
for i := 0; i < numWorkers; i++ {
<-done
}
fmt.Println("All workers done.")
Wait() // without this the defered prints from the workers doesn't flush
}
最佳答案
循环选择两个 channel 。当 channel 关闭时,将 channel 变量设置为 nil 以使该 channel 上的接收未就绪。当两个 channel 都为零时跳出循环。
http://play.golang.org/p/9gRY1yKqJ9
package main
import (
"fmt"
"time"
)
const (
fillCount = 10 // number of elements in each input channel
numWorkers = 3 // number of consumers.
)
func fillChannel(work chan string, name string) {
for i := 0; i < fillCount; i++ {
work <- fmt.Sprintf("%s%d", name, i)
}
close(work) // we're finished
}
func doWork(id int, ch1 chan string, ch2 chan string, done chan bool) {
fmt.Println("Running worker", id)
for ch1 != nil || ch2 != nil {
select {
case str, ok := <-ch1:
if ok {
fmt.Printf("%d: ch1(%d) %s\n", id, len(ch1), str)
} else {
ch1 = nil
fmt.Printf("%d: ch1 closed\n", id)
}
case str, ok := <-ch2:
if ok {
fmt.Printf("%d: ch2(%d) %s\n", id, len(ch2), str)
} else {
ch2 = nil
fmt.Printf("%d: ch2 closed\n", id)
}
}
}
fmt.Println("Ending worker", id)
done <- true
}
func main() {
a := make(chan string, 2) // a small channel
b := make(chan string, 5) // a bigger channel
// generate work
go fillChannel(a, "A")
go fillChannel(b, "B")
// launch the consumers
done := make(chan bool)
for i := 0; i < numWorkers; i++ {
go doWork(i, a, b, done)
}
// wait for the goroutines to finish.
for i := 0; i < numWorkers; i++ {
<-done
}
fmt.Println("All workers done.")
}
关于go - 如何编写更好的双 channel 选择,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25917250/