如果我们无法从传递给该例程的通道收听，如何停止 goroutine

Question

我遇到了一个关于 goroutines 的问题。假设有一个通道，我们通过来自 main 的 goroutine 传递这个通道。现在，如果我们无法从 main 收听此频道（以防在收听之前发生返回/恐慌）。goroutine 不会停止。如果出现错误，如何停止这个 goroutine？

如果多次调用 goroutine 中的函数，例程的数量会不断增加。

package main

import (

    "fmt"

    "runtime"

)

func test(a chan string) {

    defer func() {

        close(a)

        fmt.Println("channel close")

    }()

    fmt.Println("sending to channel")

    a <- "1"

    fmt.Println("sent to channel")

}

func method() string {

    fmt.Println("method starting no. of routine=>",

        runtime.NumGoroutine())

    b := make(chan string)

    go test(b)

    fmt.Println("method current no. of routine=>",

        runtime.NumGoroutine())

    return "error" //if this is executed the routines keeps on

    //increasing

    a := <-b

    return a

}

func main() {

    defer fmt.Println("final main no. of routine=>",

        runtime.NumGoroutine())

    i := 0

    //firing 10 request for method

    for {

        if i < 10 {

               fmt.Println(method())

               i++

        } else {

               break

        }

    }

}

输出：

method starting no. of routine=> 1

method current no. of routine=> 2

error

method starting no. of routine=> 2

method current no. of routine=> 3

error

method starting no. of routine=> 3

method current no. of routine=> 4

error

.....继续这样增加

Answer 1

例程可以根据上下文停止。在使用上下文之前，您应该知道只有带有循环的例程才需要停止控制，那些死例程不需要停止。上下文示例：func main(){&nbsp; &nbsp; ctx, cancel := context.WithCancel(context.Background())&nbsp; &nbsp; go func(c context.Context){&nbsp; &nbsp; &nbsp; &nbsp; for {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; select{&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; case <-c.Done():&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;fmt.Println("exit success")&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; default:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;// service&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;fmt.Println("my logic service loop")&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }(ctx)&nbsp; &nbsp; time.Sleep(5 * time.Second)&nbsp; &nbsp;cancel()}