Question

我有一个带有goroutine的程序，我们将调用mainRoutine来锁定资源，另一方面，我们将其他被触发的goroutines命名为goroutine-0 goroutine-1 goroutine-2 ....此例程尝试获取锁，在mainRoutine停止后，我需要我的其他goroutine以同步方式获取锁，我的意思是我想要goroutine-0然后goroutine-1 ... 我为解决此问题所做的工作是花费时间片。时间以goroutine启动时使用的time.Now（）填充，并使用sync.Cond。一些代码示例来说明：

package main

import (
    "fmt"
    "sync"
    "time"
)

func condition(myTime time.Time, timeSlice []time.Time) bool {
    for _, v := range timeSlice {
        if myTime.After(v) {
            return false
        }
    }
    return true
}

func removeFromSlice(myTime time.Time, timeSlice []time.Time) {
    var place int
    for i, v := range timeSlice {
        if myTime.Equal(v) {
            place = i
            break
        }
    }

    timeSlice = append(timeSlice[:place], timeSlice[place+1:]...)
}

func main() {
    var m sync.Mutex
    c := sync.NewCond(&m)

    c.L.Lock()
    fmt.Println("Locker locked")
    go func() {
        time.Sleep(time.Second * 1)

        c.L.Unlock()
        fmt.Println("Locker unlocked")
    }()

    var wg sync.WaitGroup
    var timeSlice []time.Time
    wg.Add(100)

    for i := 0; i < 100; i++ {
        now := time.Now()
        timeSlice = append(timeSlice, now)
        time.Sleep(time.Nanosecond * 1) // ensure there's at leat 1 nanosec of diff between 2 time.Now
        go func(i int, myTime time.Time) {
            fmt.Printf("Before %d %d\n", i, myTime.Unix())
            c.L.Lock()
            for !condition(myTime, timeSlice) {
                c.Wait()
            }
            c.L.Unlock()
            removeFromSlice(myTime, timeSlice)
            c.Broadcast()
            wg.Done()
            fmt.Printf("After done %d\n", i)
        }(i, now)
    }
    wg.Wait()

    fmt.Println("Hello, playground")
}

我认为这不是做这种看起来确实很棘手的事情的正确方法，还有更好的方法吗？

-编辑- 在回答@Vorsprung之后，我认为我的案例中最好的方法是制作一个函数切片，该函数总是称为该切片的第一个元素

package main

import (
    "fmt"
    "sync"
)

func makeFunc(id int) func() {
    return func() {
        fmt.Printf("called %d\n", id)
    }
}

func main() {
    var wg sync.WaitGroup
    var funcSlice []func()
    var m sync.Mutex

    for i := 0; i < 5; i++ {
        funcSlice = append(funcSlice, makeFunc(i))
        wg.Add(1)
        go func() {
            defer wg.Done()
            m.Lock()
            defer m.Unlock()
            funcSlice[0]()
            funcSlice = funcSlice[1:]
        }()
    }
    wg.Wait()
    fmt.Println("finished")
}

Answer 1

给goroutines一个内部ID，然后让它们依次调用。下面的示例说明如何工作

package main

import (
    "fmt"
    "sync"
)

func main() {
    var wg sync.WaitGroup
    var c [5]chan int
    for i := range c {
        c[i] = make(chan int)
        wg.Add(1)
        go func(id int) {
            defer wg.Done()
            f := <-c[id]
            fmt.Println("called from ", f, ".  My id ", id)
            if id < 4 {
                fmt.Println(id+1, " next")
                c[id+1] <- id
            }
            fmt.Println("ending ", id)
        }(i)
    }
    c[0] <- 99

    wg.Wait()
    fmt.Println("bye")
}

https://play.golang.org/p/psF8ISodJU_3

通过锁询问时间同步获取锁

1 个答案: