Question

在以下代码中有两个频道A＆amp; B包含工作，在实际代码中它们是不同的结构，工人需要在退出之前排空两个通道。工人需要从两个渠道进来的信息。两个选择语句有效，但它非常笨拙。如果我添加default:以使其无阻塞，则代码无法排空通道。是否有更好的方式来编写选择？

现在如果频道A没有工作，那么频道B也不会得到服务。要解决的另一个问题，但不是我的主要关注点。

playground用于测试以下代码：

package main

import (
    "fmt"
    "time"
)

const (
    fillCount  = 10 // number of elements in each input channel
    numWorkers = 3  // number of consumers.
)

func Wait() {
    time.Sleep(2000 * time.Millisecond)
}

func fillChannel(work chan string, name string) {
    for i := 0; i < fillCount; i++ {
        work <- fmt.Sprintf("%s%d", name, i)
    }
    close(work) // we're finished
}

func doWork(id int, ch1 chan string, ch2 chan string, done chan bool) {
    fmt.Println("Running worker", id)
    defer fmt.Println("Ending worker", id)

    for ch1Open, ch2Open := true, true; ch1Open && ch2Open; {
        cnt1 := len(ch1)
        cnt2 := len(ch2)

        if ch1Open {
            select {
            case str, more := <-ch1:
                if more {
                    fmt.Printf("%d: ch1(%d) %s\n", id, cnt1, str)
                } else {
                    fmt.Printf("%d: ch1 closed\n", id)
                    ch1Open = false
                }
            }
        }

        if ch2Open {
            select {
            case str, more := <-ch2:
                if more {
                    fmt.Printf("%d: ch2(%d) %s\n", id, cnt2, str)
                } else {
                    fmt.Printf("%d: ch2 closed\n", id)
                    ch2Open = false
                }
            }
        }
    }
    done <- true
}

func main() {

    a := make(chan string, 2) // a small channel
    b := make(chan string, 5) // a bigger channel

    // generate work
    go fillChannel(a, "A")
    go fillChannel(b, "B")

    // launch the consumers
    done := make(chan bool)

    for i := 0; i < numWorkers; i++ {
        go doWork(i, a, b, done)
    }

    // wait for the goroutines to finish.
    for i := 0; i < numWorkers; i++ {
        <-done
    }
    fmt.Println("All workers done.")

    Wait() // without this the defered prints from the workers doesn't flush
}

Answer 1

在循环中选择两个通道。关闭通道时，将通道变量设置为nil，以使该通道上的接收未就绪。当两个通道都为零时，跳出循环。

http://play.golang.org/p/9gRY1yKqJ9

package main

import (
    "fmt"
    "time"
)

const (
    fillCount  = 10 // number of elements in each input channel
    numWorkers = 3  // number of consumers.
)

func fillChannel(work chan string, name string) {
    for i := 0; i < fillCount; i++ {
        work <- fmt.Sprintf("%s%d", name, i)
    }
    close(work) // we're finished
}

func doWork(id int, ch1 chan string, ch2 chan string, done chan bool) {
    fmt.Println("Running worker", id)
    for ch1 != nil || ch2 != nil {
        select {
        case str, ok := <-ch1:
            if ok {
                fmt.Printf("%d: ch1(%d) %s\n", id, len(ch1), str)
            } else {
                ch1 = nil
                fmt.Printf("%d: ch1 closed\n", id)
            }

        case str, ok := <-ch2:
            if ok {
                fmt.Printf("%d: ch2(%d) %s\n", id, len(ch2), str)
            } else {
                ch2 = nil
                fmt.Printf("%d: ch2 closed\n", id)
            }

        }
    }
    fmt.Println("Ending worker", id)
    done <- true
}

func main() {

    a := make(chan string, 2) // a small channel
    b := make(chan string, 5) // a bigger channel

    // generate work
    go fillChannel(a, "A")
    go fillChannel(b, "B")

    // launch the consumers
    done := make(chan bool)

    for i := 0; i < numWorkers; i++ {
        go doWork(i, a, b, done)
    }

    // wait for the goroutines to finish.
    for i := 0; i < numWorkers; i++ {
        <-done
    }
    fmt.Println("All workers done.")
}

如何写出更好的双通道选择

1 个答案: