如何强制中断在预定滴答内花费太长时间的函数执行

Question

我已经编写了这个调度程序，但我无法"kill"当f花费超过输入recurring时间间隔时输入函数f .

如果 f 是一个进程而不是线程，那么我正在寻找的这个东西可能是某种定义的硬抢占。

f 定义是我无法控制的。它代表一个 ETL 作业，涉及在批处理执行期间处理来自多个数据库的数据。 f 它是用 go 编写的并且工作正常，但我需要以某种方式对其执行时间太长进行某种控制。

我知道 f 是原子的，所以它要么在执行结束时更改数据库，要么不更改。所以它可以被认为是安全的 "kill" 当它花费太长时间时。

func schedule(f func(), recurring time.Duration) chan struct{} {
    ticker := time.NewTicker(recurring)
    quit := make(chan struct{})
    go func(inFunc func()) {
        for {
            select {
            case <-ticker.C:
                fmt.Println("Ticked")
                // when "go" is removed, then if "f()" takes
                // more than "recurring", then it postpones
                // the following executions of "f()"
                //
                // instead somehow it should be "killed"
                // 
                // check the timestamps in the execution of the test
                go inFunc()
            case <-quit:
                fmt.Println("Stopping the scheduler")
                ticker.Stop()
                return
            }
        }
    }(f)

    return quit
}

为了看看发生了什么，我编写了这个测试：

func TestSlowExecutions(t *testing.T) {
    // log some information using a human readable timestamp
    dummyLog := func(format string, a ...interface{}) (n int, err error) {
        prefix := fmt.Sprintf("[%v] ", time.Now())
        message := fmt.Sprintf(format, a...)
        return fmt.Printf("%s%s\n", prefix, message)
    }

    // UUID to be able to uniquely identify "fooFunc"
    newUuid := func() string {
        // sudo apt-get install uuid-runtime
        uuid, _ := exec.Command("uuidgen").Output()

        re := regexp.MustCompile(`\r?\n`)
        uuidStr := re.ReplaceAllString(string(uuid), "")
        return uuidStr
    }

    // simulate some sort of very slow execution
    fooFunc := func() {
        uuid := newUuid()
        dummyLog("Ticked")
        dummyLog("Starting task %s", uuid)
        time.Sleep(2 * time.Second)
        dummyLog("Finished task %s", uuid)
    }

    // test the very slow execution of "fooFunc"
    quitChan := schedule(fooFunc, 1*time.Second)

    time.Sleep(4 * time.Second)
    close(quitChan)
    // wait more to see the "closing" message
    time.Sleep(4 * time.Second)
}

Answer 1

我与 f() 的作者协商使用超时 (https://golang.org/pkg/context/#WithTimeout) 的上下文。

请参阅下面的工作示例，注意 dummyLog 的时间戳，因此应该清楚此过程中涉及的所有 go 例程发生了什么。

代码：

// dummyLog could be used to log some information using a human readable timestamp and the benefits of `fmt.Sprintf`
func dummyLog(format string, a ...interface{}) (n int, err error) {
    prefix := fmt.Sprintf("[%v] ", time.Now())
    message := fmt.Sprintf(format, a...)
    return fmt.Printf("%s%s\n", prefix, message)
}

// newContext is providing a brand new context with a upper bound timeout
func newContext(timeoutUpperBound time.Duration) (context.Context, context.CancelFunc) {
    ctx, cancel := context.WithTimeout(context.Background(), timeoutUpperBound)
    deadline, ok := ctx.Deadline()
    dummyLog("The context deadline is set to %s is it still valid? %v", deadline, ok)
    return ctx, cancel
}

// schedule could be used to schedule arbitrary functions with a recurring interval
func schedule(f func(ctx context.Context), recurring time.Duration) chan struct{} {
    ticker := time.NewTicker(recurring)
    quit := make(chan struct{})
    go func(inFunc func(ctx context.Context)) {
        for {
            select {
            case <-ticker.C:
                dummyLog("Ticked in the scheduler")
                // simulate the "killing" of "inFunc" when it takes too long
                go func(recurring time.Duration) {
                    inCtx, cancel := newContext(recurring)
                    defer cancel()
                    inFunc(inCtx)
                }(recurring)
            case <-quit:
                dummyLog("Stopping the scheduler")
                ticker.Stop()
                return
            }
        }
    }(f)

    return quit
}

在测试环境中执行代码（虽然没有执行断言）：

func TestSomething(t *testing.T) {

    // newUuid could be used to generate a UUID to be able to uniquely identify "fooFunc"
    newUuid := func() string {
        // sudo apt-get install uuid-runtime
        uuid, _ := exec.Command("uuidgen").Output()

        re := regexp.MustCompile(`\r?\n`)
        uuidStr := re.ReplaceAllString(string(uuid), "")
        return uuidStr
    }

    // randBetween is a dummy random int generator using "math/rand"
    randBetween := func(min int, max int) int {
        return min + rand.Intn(max-min)
    }

    // fooFunc simulates some sort of very slow execution
    // like database queries or network I/O
    fooFunc := func(ctx context.Context) {
        uuid := newUuid()
        randWait := time.Duration(randBetween(0, 4000)) * time.Millisecond
        dummyLog("Starting task %s taking %s random time", uuid, randWait)
        select {
        case <-time.After(randWait):
            dummyLog("Finished task %s", uuid)
        case <-ctx.Done():
            dummyLog("Killed task %s, reason: '%s'", uuid, ctx.Err())
        }
    }

    // test the very slow execution of "fooFunc"
    timeoutUpperBound := 2 * time.Second
    quitChan := schedule(fooFunc, timeoutUpperBound)

    time.Sleep(6 * timeoutUpperBound)
    close(quitChan)
    // wait more to see the "closing" message
    time.Sleep(4 * time.Second)
}