英文:
it is quite common way to implement a timeout, why time.After is not working
问题
以下是翻译好的内容：
下面是Go代码，我尝试通过Go协程打印"result"。为什么在3秒后没有打印"timeout"，而是在10秒后打印。因为所有的10个"result"都被打印了。
package mainimport "fmt"import "time"func main() {ch := make(chan string)go func() {for i:=0;i<10;i++ {time.Sleep(time.Second * 1)ch <- "result"}}()for {select {case res := <-ch:fmt.Println(res)case <-time.After(time.Second * 3):fmt.Println("timeout")return}}}/**resultresultresultresultresultresultresultresultresultresulttimeout*/
英文:
the go code is show below, i try to print "result" by go routine. why "timeout" is not print after 3 second, but 10 second.since all the 10 "result" print.
package mainimport "fmt"import "time"func main() {ch := make(chan string)go func() {for i:=0;i<10;i++ {time.Sleep(time.Second * 1)ch <- "result"}}()for {select {case res := <-ch:fmt.Println(res)case <-time.After(time.Second * 3):fmt.Println("timeout")return}}}/**resultresultresultresultresultresultresultresultresultresulttimeout*/
答案1
得分: 3
在每个for循环中，在case res := <-ch之后，你创建了一个新的channel <-time.After(time.Second * 3)。
这个修复可能是你想要的：
timeoutChan := time.After(time.Second * 3)for {select {case res := <-ch:fmt.Println(res)case <-timeoutChan:fmt.Println("timeout")return}}
英文:
In each for loop, after case res := <-ch, you create a new channel <-time.After(time.Second * 3).
This fix maybe what you want:
timeoutChan := time.After(time.Second * 3)for {select {case res := <-ch:fmt.Println(res)case <-timeoutChan:fmt.Println("timeout")return}}
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