去哪儿面试题（Golang实现）多线程顺序输出1-75

FredricZhu ·

package main

import (
    "fmt"
    "sync"
    "time"
)

var waiter sync.WaitGroup

func print5(i int) {
    j := i
    for j < i+5 {
        fmt.Println(j)
        j++
    }
    waiter.Done()
}

func main() {
    fmt.Println("Start print 75")
    waiter = sync.WaitGroup{}
    waiter.Add(15)

    for i := 1; i < 75; i += 5 {
        go print5(i)
        time.Sleep(1 * time.Nanosecond)
    }

    waiter.Wait()
    fmt.Println("End print 75")
}

程序输出如下，

image.png

本文来自：简书

感谢作者：FredricZhu

查看原文：去哪儿面试题（Golang实现）多线程顺序输出1-75

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etog

给他上个链表 :dog: ```golang package main import ( "fmt" "time" ) type null struct{} func main() { n := 75 current := make(chan null) next := make(chan null) head := current for i := 0; i < n; i++ { go print(i, current, next) current = next next = make(chan null) } close(head) time.Sleep(time.Second) } func print(i int, r, s chan null) { _, ok := <-r if !ok { fmt.Println(i + 1) if s != nil { close(s) } } } ```

评论于 2022-07-11 15:37:23

#11

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ska-guilinstyle

``` var a struct { Mu sync.RWMutex Count int } var wg sync.WaitGroup func main() { wg = sync.WaitGroup{} wg.Add(75) for i:=1; i<=75; i++ { go Add() } wg.Wait() } func Add() { defer wg.Done() a.Mu.Lock() a.Count++ fmt.Println(a.Count) a.Mu.Unlock() } ```

评论于 2019-06-07 00:07:54

#1