等待并发工作人员完成后再退出

Question

下面的代码有一个明显的问题：程序会在工人完成所有工作之前退出。在发送者开始发送数据之前，worker 的 Goroutines 被启动，数据必须保留。从 sender 函数启动这些 goroutine 不是一种选择。这样做很容易，但是需要学习更复杂的同步技术。等待工人完成的正确方法是什么？尝试关闭worker1CH和worker2CH通道，以及为每个工作人员添加专用的 sync.WaitGroups。package mainimport (    "log"    "math/rand"    "sync")func main() {    worker1CH := make(chan int, 1)    worker2CH := make(chan int, 1)    // worker for even numbers    go func(in chan int) {        for i := range in {            log.Print(i)        }    }(worker1CH)    // worker for odd numbers    go func(in chan int) {        for i := range in {            log.Print(i)        }    }(worker2CH)    // sender which sends even numbers to worker1CH, and odd numbers to worker2CH    var wg sync.WaitGroup    wg.Add(1)    go func(wg *sync.WaitGroup, evenChan chan int, oddChan chan int) {        defer wg.Done()        data := rand.Perm(10)        for _, i := range data {            switch i%2 {            case 0:                evenChan <- i            default:                oddChan <- i            }        }    }(&wg, worker1CH, worker2CH)    wg.Wait()}

Answer 1

使用等待组等待两个接收 goroutine 完成。使用一个等待组来等待两个 goroutine。

发送完所有值后关闭通道，以便接收 goroutine 中的循环退出。

无需等待发送 goroutine。在其他协程完成之前，灌浆会完成所有工作。

worker1CH := make(chan int, 1)

worker2CH := make(chan int, 1)

var wg sync.WaitGroup

wg.Add(2)  // <-- wait for the two receiving goroutines.

// worker for even numbers

go func(wg *sync.WaitGroup, in chan int) {

    defer wg.Done() // <--- add this line

    for i := range in {

        log.Print(i)

    }

}(&wg, worker1CH)

// worker for odd numbers

go func(wg *sync.WaitGroup, in chan int) {

    defer wg.Done() <-- add this line

    for i := range in {

        log.Print(i)

    }

}(&wg, worker2CH)

// sender which sends even numbers to worker1CH, and odd numbers to worker2CH

go func(evenChan chan int, oddChan chan int) {

    defer close(evenChan) // <-- close channel so that receiver exits loop

    defer close(oddChan)  // <-- ditto

    data := rand.Perm(10)

    for _, i := range data {

        switch i % 2 {

        case 0:

            evenChan <- i

        default:

            oddChan <- i

        }

    }

}(worker1CH, worker2CH)

wg.Wait()

Answer 2

已经能够创建worker1Done和worker2Done通道，然后等待工作完成。

还必须将 close(evenChan) 和 close(oddChan) 添加到 sender 函数以避免fatal error: all goroutines are asleep - deadlock!错误

package main

import (

    "log"

    "math/rand"

    "sync"

)

func main() {

    worker1CH := make(chan int, 1)

    worker2CH := make(chan int, 1)

    worker1Done := make(chan bool)

    worker2Done := make(chan bool)

    // worker for even numbers

    go func(in chan int, done chan bool) {

        for i := range in {

            log.Print(i)

        }

        done <- true

    }(worker1CH, worker1Done)

    // worker for odd numbers

    go func(in chan int, done chan bool) {

        for i := range in {

            log.Print(i)

        }

        done <- true

    }(worker2CH, worker2Done)

    // sender which sends even numbers to worker1CH, and odd numbers to worker2CH

    var wg sync.WaitGroup

    wg.Add(1)

    go func(wg *sync.WaitGroup, evenChan chan int, oddChan chan int) {

        defer wg.Done()

        data := rand.Perm(10)

        for _, i := range data {

            switch i%2 {

            case 0:

                evenChan <- i

            default:

                oddChan <- i

            }

        }

        close(evenChan)

        close(oddChan)

    }(&wg, worker1CH, worker2CH)

    wg.Wait()

    <- worker1Done

    <- worker2Done

}

Answer 3

由于您的发件人具有固定大小，因此它将自行退出，您可以关闭阅读器通道并等待

package main

import (

    "log"

    "math/rand"

    "sync"

)

func reader(in chan int, wg *sync.WaitGroup) {

    defer wg.Done()

    for i := range in {

        log.Print(i)

    }

}

func main() {

    var wg sync.WaitGroup

    worker1CH := make(chan int, 1)

    worker2CH := make(chan int, 1)

    wg.Add(1)

    // worker for even numbers

    go reader(worker1CH, &wg)

    wg.Add(1)

    // worker for odd numbers

    go reader(worker2CH, &wg)

    // sender which sends even numbers to worker1CH, and odd numbers to worker2CH

    sender(worker1CH, worker2CH)

    close(worker2CH)

    close(worker1CH)

    wg.Wait()

}

func sender(evenChan chan int, oddChan chan int) {

    data := rand.Perm(10)

    for _, i := range data {

        switch i % 2 {

        case 0:

            evenChan <- i

        default:

            oddChan <- i

        }

    }

}

游乐场链接https://play.golang.org/p/JJ9ngCHUvbS