如何在执行一组工人之间适当地延迟

Question

美好的一天,

我正在尝试在工作人员执行之间实现正确的延迟，例如，工作人员必须完成 30 个任务并休眠 5 秒，我如何在代码中准确跟踪30 个任务 已完成，仅在完成后才进入休眠 5 秒?

下面是创建 30 名工人 池的代码，这些工人依次以无序方式一次执行 30 件任务，代码如下：


import (
    "fmt"
    "math/rand"
    "sync"
    "time"
)

type Job struct {
    id       int
    randomno int
}
type Result struct {
    job         Job
    sumofdigits int
}

var jobs = make(chan Job, 10)
var results = make(chan Result, 10)

func digits(number int) int {
    sum := 0
    no := number
    for no != 0 {
        digit := no % 10
        sum += digit
        no /= 10
    }
    time.Sleep(2 * time.Second)
    return sum
}
func worker(wg *sync.WaitGroup) {
    for job := range jobs {
        output := Result{job, digits(job.randomno)}
        results <- output
    }
    wg.Done()
}
func createWorkerPool(noOfWorkers int) {
    var wg sync.WaitGroup
    for i := 0; i < noOfWorkers; i++ {
        wg.Add(1)
        go worker(&wg)
    }

    wg.Wait()
    close(results)
}
func allocate(noOfJobs int) {
    for i := 0; i < noOfJobs; i++ {
        if i != 0 && i%30 == 0 {
            fmt.Printf("SLEEPAGE 5 sec...")
            time.Sleep(10 * time.Second)

        }
        randomno := rand.Intn(999)
        job := Job{i, randomno}
        jobs <- job
    }
    close(jobs)
}
func result(done chan bool) {
    for result := range results {
        fmt.Printf("Job id %d, input random no %d , sum of digits %d\n", result.job.id, result.job.randomno, result.sumofdigits)
    }
    done <- true
}
func main() {
    startTime := time.Now()
    noOfJobs := 100
    go allocate(noOfJobs)
    done := make(chan bool)
    go result(done)
    noOfWorkers := 30
    createWorkerPool(noOfWorkers)
    <-done
    endTime := time.Now()
    diff := endTime.Sub(startTime)
    fmt.Println("total time taken ", diff.Seconds(), "seconds")
}

播放：https://go.dev/play/p/lehl7hoo-kp

具体如何确保30个任务完成以及在何处插入延迟尚不清楚，如有帮助，将不胜感激

Answer 1

好的，让我们从这个工作示例开始：

func Test_t(t *testing.T) {

    // just a published, this publishes result on a chan
    publish := func(s int, ch chan int, wg *sync.WaitGroup) {
        ch <- s // this is blocking!!!
        wg.Done()
    }

    wg := &sync.WaitGroup{}
    wg.Add(100)

    // we'll use done channel to notify the work is done
    res := make(chan int)
    done := make(chan struct{})
    // create worker that will notify that all results were published
    go func() {
        wg.Wait()
        done <- struct{}{}
    }()
    
    // let's create a jobs that publish on our res chan
    // please note all goroutines are created immediately
    for i := 0; i < 100; i++ {
        go publish(i, res, wg)
    }

    // lets get 30 args and then wait
    var resCounter int
forloop:
    for {
        select {
        case ss := <-res:
            println(ss)
            resCounter += 1
            // break the loop
            if resCounter%30 == 0 {
                // after receiving 30 results we are blocking this thread
                // no more results will be taken from the channel for 5 seconds
                println("received 30 results, waiting...")
                time.Sleep(5 * time.Second)
            }
        case <-done:
            // we are done here, let's break this infinite loop
            break forloop
        }
    }
}

我希望这能说明如何做到这一点。

那么，您的代码有什么问题？老实说，看起来不错（我的意思是发布了 30 个结果，然后是代码等待，然后是另外 30 个结果，等等），但问题是 您想在哪里等待？

我猜有几种可能：

创建工人（这就是您的代码现在的工作方式，如我所见，它以 30 包的形式发布工作；请注意 digit 中的 2 秒延迟函数仅适用于执行代码的goroutine)
触发工人（所以“等待”代码应该在工人函数中，不允许运行更多工人 - 所以它必须观察有多少结果被发布）
处理结果（这就是我的代码的工作方式，forloop 中的正确同步）

如何在执行一组工人之间适当地延迟

How to properly delay between executing a pool of workers

concurrency

go

goroutine