等待并发工作人员完成后再退出

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下面的代码有一个明显的问题：程序会在worker完成所有工作之前退出。

工作线程的 goroutines 在发送者开始发送数据之前启动，数据必须保留。从发送者函数启动这些 goroutine 不是一个选项。这样做很容易，但是需要学习更复杂的同步技术。

等待工作人员完成的正确方法是什么？

已尝试关闭 worker1ch 和 worker2ch 通道，以及向每个工作线程添加专用的sync.waitgroups。

package main

import (
    "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()

}

解决方案

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()

Run the example on the Go Playground。

已经能够创建 worker1done 和 worker2done 通道，然后等待工作完成。

还必须向发送者函数添加 close(evenchan) 和 close(oddchan) 以避免 fatal 错误：所有 goroutine 都在睡眠 - 死锁！ 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

}

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