Question

我正在寻找一种可靠的方法来确保Go中的空信道不会阻止我的执行。我必须以特定顺序（优先级）迭代多个频道，一旦我找到一个包含项目的频道，请阅读一个。

目前我以类似的方式做事：

if len(myChannel) > 0 {
    // Possible issue here: length could have changed to 0 making this blocking
    elm := <- myChannel
    return elm
}

从理论上讲，这可能会导致等待时间过长，而另一个频道可能会有一个准备好服务的项目＆＃34;。

有关如何改进的任何建议？我可以在频道中使用互斥锁，但感觉这是一个更好的解决方案，虽然我不确定如何。

Answer 1

有reflect.Select函数可以执行您想要的操作：

package main

import (
    "fmt"
    "reflect"
    "time"
)

func main() {
    a, b, c := make(chan int), make(chan int), make(chan int)
    go func() {
        time.Sleep(2 * time.Second)
        a <- 1
    }()
    go func() {
        time.Sleep(time.Second)
        b <- 2
    }()
    go func() {
        time.Sleep(3 * time.Second)
        c <- 3
    }()
    for i := 0; i < 3; i++ {
        chosen, recv, ok := reflect.Select([]reflect.SelectCase{
            reflect.SelectCase{
                Dir:  reflect.SelectRecv,
                Chan: reflect.ValueOf(a),
            },
            reflect.SelectCase{
                Dir:  reflect.SelectRecv,
                Chan: reflect.ValueOf(b),
            },
            reflect.SelectCase{
                Dir:  reflect.SelectRecv,
                Chan: reflect.ValueOf(c),
            },
        })
        if ok {
            fmt.Printf("Got value %d from %d\n", recv.Interface().(int), chosen)
        }
    }
}

play.golang.org

Answer 2

I'm not sure this really answers the question "Is there a reliable way to ensure a Go channel does not block". In the OP's use case, it's OK if a recv blocks /so long as no other channel in the set would not block/, and that's what the accepted solution implements. This is different from "ensure the recv does not block", and I see no way around the fundamental limitation of the race condition OP points out.

确保Go通道不会阻塞的可靠方法

2 个答案: