一尘不染

GO语言:致命错误:所有goroutine都在睡眠中-死锁

go

下面的代码与硬编码的JSON数据一起正常工作,但是当我从文件中读取JSON数据时不起作用。我得到fatal error: all goroutines are asleep - deadlock使用时的错误sync.WaitGroup

硬编码JSON数据的工作示例:

package main

import (
    "bytes"
    "fmt"
    "os/exec"
    "time"
)

func connect(host string) {
    cmd := exec.Command("ssh", host, "uptime")
    var out bytes.Buffer
    cmd.Stdout = &out
    err := cmd.Run()
    if err != nil {
        fmt.Println(err)
    }
    fmt.Printf("%s: %q\n", host, out.String())
    time.Sleep(time.Second * 2)
    fmt.Printf("%s: DONE\n", host)
}

func listener(c chan string) {
    for {
        host := <-c
        go connect(host)
    }
}

func main() {
    hosts := [2]string{"user1@111.79.154.111", "user2@111.79.190.222"}
    var c chan string = make(chan string)
    go listener(c)

    for i := 0; i < len(hosts); i++ {
        c <- hosts[i]
    }
    var input string
    fmt.Scanln(&input)
}

输出:

user@user-VirtualBox:~/go$ go run channel.go
user1@111.79.154.111: " 09:46:40 up 86 days, 18:16,  0 users,  load average: 5"
user2@111.79.190.222: " 09:46:40 up 86 days, 17:27,  1 user,  load average: 9"
user1@111.79.154.111: DONE
user2@111.79.190.222: DONE

不起作用-读取JSON数据文件的示例:

package main

import (
    "bytes"
    "fmt"
    "os/exec"
    "time"
    "encoding/json"
    "os"
    "sync"
)

func connect(host string) {
    cmd := exec.Command("ssh", host, "uptime")
    var out bytes.Buffer
    cmd.Stdout = &out
    err := cmd.Run()
    if err != nil {
        fmt.Println(err)
    }
    fmt.Printf("%s: %q\n", host, out.String())
    time.Sleep(time.Second * 2)
    fmt.Printf("%s: DONE\n", host)
}

func listener(c chan string) {
    for {
        host := <-c
        go connect(host)
    }
}

type Content struct {
    Username string `json:"username"`
    Ip       string `json:"ip"`
}

func main() {
    var wg sync.WaitGroup

    var source []Content
    var hosts []string
    data := json.NewDecoder(os.Stdin)
    data.Decode(&source)

    for _, value := range source {
        hosts = append(hosts, value.Username + "@" + value.Ip)
    }

    var c chan string = make(chan string)
    go listener(c)

    for i := 0; i < len(hosts); i++ {
        wg.Add(1)
        c <- hosts[i]
        defer wg.Done()
    }

    var input string
    fmt.Scanln(&input)

    wg.Wait()
}

输出值

user@user-VirtualBox:~/go$ go run deploy.go < hosts.txt 
user1@111.79.154.111: " 09:46:40 up 86 days, 18:16,  0 users,  load average: 5"
user2@111.79.190.222: " 09:46:40 up 86 days, 17:27,  1 user,  load average: 9"
user1@111.79.154.111 : DONE
user2@111.79.190.222: DONE
fatal error: all goroutines are asleep - deadlock!

goroutine 1 [semacquire]:
sync.runtime_Semacquire(0xc210000068)
    /usr/lib/go/src/pkg/runtime/sema.goc:199 +0x30
sync.(*WaitGroup).Wait(0xc210047020)
    /usr/lib/go/src/pkg/sync/waitgroup.go:127 +0x14b
main.main()
    /home/user/go/deploy.go:64 +0x45a

goroutine 3 [chan receive]:
main.listener(0xc210038060)
    /home/user/go/deploy.go:28 +0x30
created by main.main
    /home/user/go/deploy.go:53 +0x30b
exit status 2
user@user-VirtualBox:~/go$

主机

[
   {
      "username":"user1",
      "ip":"111.79.154.111"
   },
   {
      "username":"user2",
      "ip":"111.79.190.222"
   }
]

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2020-07-02

共1个答案

一尘不染

当主功能结束时,Go程序结束。

语言规范

程序执行首先初始化主程序包,然后调用函数main。当该函数调用返回时,程序退出。它不等待其他(非主)goroutine完成。

因此,您需要等待goroutine完成。常见的解决方案是使用sync.WaitGroup对象。

最简单的同步goroutine的代码:

package main

import "fmt"
import "sync"

var wg sync.WaitGroup // 1

func routine() {
    defer wg.Done() // 3
    fmt.Println("routine finished")
}

func main() {
    wg.Add(1) // 2
    go routine() // *
    wg.Wait() // 4
    fmt.Println("main finished")
}

并用于同步多个goroutine

package main

import "fmt"
import "sync"

var wg sync.WaitGroup // 1

func routine(i int) {
    defer wg.Done() // 3
    fmt.Printf("routine %v finished\n", i)
}

func main() {
    for i := 0; i < 10; i++ {
        wg.Add(1) // 2
        go routine(i) // *
    }
    wg.Wait() // 4
    fmt.Println("main finished")
}

WaitGroup按执行顺序使用。

  1. 声明全局变量。使它全局化是使其对所有功能和方法可见的最简单方法。
  2. 增加柜台。这必须在主goroutine中完成,因为由于内存模型的保证,不能保证新启动的goroutine将在4之前执行。
  3. 减少计数器。这必须在goroutine的出口处完成。使用延迟调用,我们确保无论函数结束如何结束,只要函数结束,都将调用它。
  4. 等待计数器达到0。必须在主goroutine中执行此操作,以防止程序退出。

*实际参数在开始新的鱼肉素之前进行评估。因此,需要在对它们进行显式评估之前,wg.Add(1)使可能出现恐慌的代码不会留下增加的计数器。

param := f(x)
wg.Add(1)
go g(param)

代替

wg.Add(1)
go g(f(x))
2020-07-02