我正在嘗試理解 Golang 例程的簡單代碼：package mainimport ( "fmt" "time")func sleep(seconds int, endSignal chan<- bool) { time.Sleep(time.Duration(seconds) * time.Second) endSignal <- true}func main() { endSignal := make(chan bool, 1) go sleep(3, endSignal) var end bool for !end { select { case end = <-endSignal: fmt.Println("The end!") case <-time.After(5 * time.Second): fmt.Println("There's no more time to this. Exiting!") end = true } }}很好，但是為什么我不能在這個“選擇”塊中使用簡單的默認值？像這樣的東西：for !end { select { case end = <-endSignal: fmt.Println("The end.") case <-time.After(4 * time.Second): fmt.Println("There's no more time to this. Exiting!") end = true default: fmt.Println("No end signal received.") }}它得到這個輸出：? go run goroutines-timeout.goNo end signal received!No end signal received!No end signal received!No end signal received!...The end!我不明白為什么。
1 回答
慕斯王
TA貢獻1864條經驗 獲得超2個贊
每次執行時，time.After(4 * time.Second)您都會創建一個新的計時器通道。該select語句無法記住它在上一次迭代中選擇的通道。您還采用了異步操作并將其變成了一個繁忙的循環，從而違背了select語句的目的。
您所需要的只是圍繞您感興趣的兩個頻道進行簡單的選擇。它根本不需要循環。
select {
case <-endSignal:
fmt.Println("The end!")
case <-time.After(4 * time.Second):
fmt.Println("There's no more time to this. Exiting!")
}
https://play.golang.org/p/jb4EE8e6cw
如果您真的想多次輪詢，請將計時器設置在 for 循環之外，以便每次迭代都檢查相同的計時器
timeout := time.After(5 * time.Second)
pollInt := time.Second
for {
select {
case <-endSignal:
fmt.Println("The end!")
return
case <-timeout:
fmt.Println("There's no more time to this. Exiting!")
return
default:
fmt.Println("still waiting")
}
time.Sleep(pollInt)
}
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