Question

我遇到了一个goroutines问题。假设有一个通道，我们通过main上的goroutine通过了该通道。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。出现错误时如何停止此goroutine？

如果在goroutine中多次调用该函数，则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

.....这样的增长趋势不断

Answer 1

例程可以根据上下文停止。在使用上下文之前，应该知道只有带循环的例程才是期望的停止控制，那些可破坏的例程无需停止。

上下文示例：

func main(){
    ctx, cancel := context.WithCancel(context.Background())
    go func(c context.Context){
        for {
            select{
                case <-c.Done():
                   fmt.Println("exit success")
                default:
                   // service
                   fmt.Println("my logic service loop")
            }    
        }
    }(ctx)
    time.Sleep(5 * time.Second)
   cancel()
}

如果我们无法通过传递给该例程的频道进行监听，如何停止goroutine

1 个答案: