英文:
Trying to understand go func with WaitGroup
问题
我有以下代码 https://play.golang.org/p/9jPlypO4d-
package main
import (
"fmt"
"sync"
"time"
)
func main() {
var wg sync.WaitGroup
wg.Add(1)
c := make(chan int)
go func() {
defer wg.Done()
for {
if <-c == -1 {
fmt.Print(":")
return
}
fmt.Print(".")
time.Sleep(time.Second)
}
}()
c <- 0
time.Sleep(time.Second * 5)
c <- -1
wg.Wait()
}
我想知道为什么只打印了一个 .？难道不应该是4或5个吗？
英文:
I have the following code https://play.golang.org/p/9jPlypO4d-
package main
import (
"fmt"
"sync"
"time"
)
func main() {
var wg sync.WaitGroup
wg.Add(1)
c := make(chan int)
go func() {
defer wg.Done()
for {
if <-c == -1 {
fmt.Print(":")
return
}
fmt.Print(".")
time.Sleep(time.Second)
}
}()
c <- 0
time.Sleep(time.Second * 5)
c <- -1
wg.Wait()
}
I wonder why there is only one . printed? Shouldn't it be like 4 or 5?
答案1
得分: 3
if <-c == -1 会阻塞直到通道中有内容。所以，第一个值是0，它获取到了这个值，打印出一个.，休眠一秒钟（而在goroutine外部休眠5秒钟），然后阻塞直到获取到下一个值。然后它返回。
我所知道的唯一不阻塞读取通道的方法是使用带有默认情况的select语句。
go func() {
defer wg.Done()
for {
select {
case x, ok := <-c:
if ok && x == -1 {
fmt.Print(":")
return
}
default:
fmt.Print(".")
}
time.Sleep(time.Second)
}
}()
链接：https://play.golang.org/p/nOG_hfih4D
英文:
if <-c == -1 will block until there is something in the channel. So, the first value is 0, it gets it, prints out a ., sleeps one second (while outside the goroutine it is sleeping 5 seconds), then it blocks until it gets the next value. And then it returns.
The only way to read a channel without blocking (that I know of) is to use a select statement with a default case.
go func() {
defer wg.Done()
for {
select {
case x, ok := <-c:
if ok && x == -1 {
fmt.Print(":")
return
}
default:
fmt.Print(".")
}
time.Sleep(time.Second)
}
}()
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