draxu26480 2014-07-22 13:57
浏览 162
已采纳

为什么睡眠在goroutine中不起作用

package main

import (
    "fmt"
    "time"
)

func main() {
    c := make(chan struct{})
    count := 1
    go func() {
        for {
            fmt.Println("foo", count)
            count++
            time.Sleep(2)
        }
        c <- struct{}{}
    }()
    fmt.Println("Hello World!")
    <-c
}

This is my code , and I found it didn't sleep 2 every loop, and printed quickly. What's the reason of it? What I searched is that sleep will make goroutine give up control of cpu and when it get the control again will it check itself is sleeping?

  • 写回答

1条回答 默认 最新

  • dth20986 2014-07-22 14:02
    关注

    time.Sleep takes its Duration in nanoseconds, so to delay 2 seconds it should be;

    time.Sleep(2000000000)    
    

    or, as @Ainar-G points out in the comments, the more readable;

    time.Sleep(2 * time.Second)
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?

悬赏问题

  • ¥15 请问有会的吗,用MATLAB做
  • ¥15 phython如何实现以下功能?查找同一用户名的消费金额合并—
  • ¥15 ARIMA模型时间序列预测用pathon解决
  • ¥15 孟德尔随机化怎样画共定位分析图
  • ¥18 模拟电路问题解答有偿速度
  • ¥15 CST仿真别人的模型结果仿真结果S参数完全不对
  • ¥15 误删注册表文件致win10无法开启
  • ¥15 请问在阿里云服务器中怎么利用数据库制作网站
  • ¥60 ESP32怎么烧录自启动程序,怎么查看客户esp32板子上程序及烧录地址
  • ¥50 html2canvas超出滚动条不显示