如何让Go通道工作者有不同的结果长度？

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我对 gobyexample 进行了一些编辑：

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

type demoresult struct {
    name string
    rate int
}

func random(min, max int) int {
    rand.seed(time.now().utc().unixnano())
    return rand.intn(max-min) + min
}

func worker(id int, jobs <-chan int, results chan<- demoresult) {
    for j := range jobs {
        fmt.println("worker", id, "started  job", j)
        time.sleep(time.second)
        fmt.println("worker", id, "finished job", j)
        myrand := random(1, 4)
        if myrand == 2 {
            results <- demoresult{name: "succ", rate: j}
        }
        //  else {
        //  results <- demoresult{name: "failed", rate: 999}
        // }
    }
}

func main() {
    const numjobs = 5
    jobs := make(chan int, numjobs)
    results := make(chan demoresult)
    for w := 1; w <= 3; w++ {
        go worker(w, jobs, results)
    }
    for j := 1; j <= numjobs; j++ {
        jobs <- j
    }
    close(jobs)

    for a := 1; a <= numjobs; a++ {
        out := <-results
        if out.name == "succ" {
            fmt.printf("%v\n", out)
        }
    }
}

我故意对以下代码进行了注释，使其永远卡住：

        //  else {
        //  results <- DemoResult{Name: "failed", Rate: 999}
        // }

看来我们应该使结果的长度与 jobs 的长度相同。我想知道我们是否可以使其具有不同的长度？

正确答案

使用 wait group 检测工作人员何时完成。当工作人员完成后，关闭结果通道。接收结果直到通道关闭。

func worker(wg *sync.waitgroup, id int, 
            jobs <-chan int, 
            results chan<- demoresult) {
    // decrement wait group counter on return from
    // function.
    defer wg.done()
    ⋮ 
}


func main() {
    ⋮
    // declare wait group and increment counter for
    // each worker.
    var wg sync.waitgroup
    for w := 1; w <= 3; w++ {
        wg.add(1)
        go worker(&wg, w, jobs, results)
    }
    ⋮
    // wait for workers to decrement wait group
    // counter to zero and close channel.
    // execute in goroutine so we can continue on 
    // to receiving values from results in main. 
    go func() {
        wg.wait()
        close(results)
    }()
    ⋮
    // loop until results is closed.
    for out := range results {
       ⋮
    }
}

https://go.dev/play/p/FOQwybMl7tM

当然可以，但是您需要某种方法来确定何时到达结果末尾。这就是您的示例失败的原因 - 当前该函数假设将有 numjobs （每个作业一个结果）结果并等待那么多结果。

另一种方法是使用通道关闭来指示这一点，即 (playground)

package main

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

type DemoResult struct {
    Name string
    Rate int
}

func random(min, max int) int {
    rand.Seed(time.Now().UTC().UnixNano())
    return rand.Intn(max-min) + min
}

func worker(id int, jobs <-chan int, results chan<- DemoResult) {
    for j := range jobs {
        fmt.Println("worker", id, "started  job", j)
        time.Sleep(time.Second)
        fmt.Println("worker", id, "finished job", j)
        myrand := random(1, 4)
        if myrand == 2 {
            results <- DemoResult{Name: "succ", Rate: j}
        } // else {
        //  results <- DemoResult{Name: "failed", Rate: 999}
        //}
    }
}

func main() {
    const numWorkers = 3
    const numJobs = 5
    jobs := make(chan int, numJobs)
    results := make(chan DemoResult)

    var wg sync.WaitGroup
    wg.Add(numWorkers)
    for w := 1; w <= numWorkers; w++ {
        go func() {
            worker(w, jobs, results)
            wg.Done()
        }()
    }
    go func() {
        wg.Wait() // Wait for go routines to complete then close results channel
        close(results)
    }()

    for j := 1; j <= numJobs; j++ {
        jobs <- j
    }
    close(jobs)

    for out := range results {
        if out.Name == "succ" {
            fmt.Printf("%v\n", out)
        }
    }
}

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