Question

在我的代码中有三个并发例程。我试着简要介绍一下我的代码，

Routine 1 {
do something

*Send int to Routine 2
Send int to Routine 3
Print Something
Print Something*

do something
}

Routine 2 {
do something

*Send int to Routine 1
Send int to Routine 3
Print Something
Print Something*

do something
}

Routine 3 {
do something

*Send int to Routine 1
Send int to Routine 2
Print Something
Print Something*

do something
}

main {
routine1
routine2
routine3
}

我想要的是，当两个代码之间的代码执行时（两个星号之间的代码）正在执行，控制流不能转到其他goroutines。例如，当routine1正在执行两颗星之间的事件（发送和打印事件）时，必须阻止例程2和3（意味着执行流程不会从例程1传递到例程2或3）。完成最后一次打印事件后，执行流程可能会传递到例程2或3.任何人都可以通过指定帮助我，我该如何实现？是否可以通过WaitGroup实现上述规范？任何人都可以通过一个简单的例子向我展示如何使用WaitGroup实现上面指定的例子。感谢。

注意：我提供了两个发送和两个打印选项，实际上有很多发送和打印。

Answer 1

如果我弄错了，你想要的是防止同时执行每个功能的某些部分和其他功能。以下代码执行此操作：fmt.Println行不会在其他例程运行时发生。这是发生的事情：当执行到达打印部分时，它等待直到其他例程结束，如果它们正在运行，并且当该打印行正在执行时，其他例程不会启动并等待。我希望这就是你要找的。如果我对此错了，请纠正我。

package main

import (
    "fmt"
    "rand"
    "sync"
)

var (
    mutex1, mutex2, mutex3 sync.Mutex
    wg sync.WaitGroup
)

func Routine1() {
    mutex1.Lock()
    // do something
    for i := 0; i < 200; i++ {
        mutex2.Lock()
        mutex3.Lock()
        fmt.Println("value of z")
        mutex2.Unlock()
        mutex3.Unlock()
    }
    // do something
    mutex1.Unlock()
    wg.Done()
}

func Routine2() {
    mutex2.Lock()
    // do something
    for i := 0; i < 200; i++ {
        mutex1.Lock()
        mutex3.Lock()
        fmt.Println("value of z")
        mutex1.Unlock()
        mutex3.Unlock()
    }
    // do something
    mutex2.Unlock()
    wg.Done()
}

func Routine3() {
    mutex3.Lock()
    // do something
    for i := 0; i < 200; i++ {
        mutex1.Lock()
        mutex2.Lock()
        fmt.Println("value of z")
        mutex1.Unlock()
        mutex2.Unlock()
    }
    // do something
    mutex3.Unlock()
    wg.Done()
}

func main() {
    wg.Add(3)
    go Routine1()
    go Routine2()
    Routine3()
    wg.Wait()
}

更新：让我在这里解释这三个互斥锁：互斥锁是documentation says：“互斥锁定”。这意味着当您在互斥锁上调用Lock时，如果其他人锁定了相同的互斥锁，您的代码就会等待。在您致电Unlock之后，阻止的代码将会恢复。

这里我将每个函数放在自己的互斥锁中，方法是在函数开头锁定一个互斥锁，然后将其解锁。通过这种简单的机制，您可以避免在运行这些函数的同时运行所需的任何代码部分。例如，在Routine1运行时，您希望拥有一个不应该运行的代码，只需在该代码的开头锁定mutex1，最后解锁即可。这就是我在Routine2和Routine3中的相应行中所做的。希望澄清一切。

Answer 2

您可以使用sync.Mutex。例如，im.Lock()和im.Unlock()之间的所有内容都将排除其他goroutine。

package main

import (
"fmt"
"sync"
)

func f1(wg *sync.WaitGroup, im *sync.Mutex, i *int) {
  for {
    im.Lock()
    (*i)++
    fmt.Printf("Go1: %d\n", *i)
    done := *i >= 10
    im.Unlock()
    if done {
      break
    }
  }
  wg.Done()
}

func f2(wg *sync.WaitGroup, im *sync.Mutex, i *int) {
  for {
    im.Lock()
    (*i)++
    fmt.Printf("Go2: %d\n", *i)
    done := *i >= 10
    im.Unlock()
    if done {
      break
    }
  }
  wg.Done()
}

func main() {
    var wg sync.WaitGroup

    var im sync.Mutex
    var i int

    wg.Add(2)
    go f1(&wg, &im, &i)
    go f2(&wg, &im, &i)
    wg.Wait()   

}

Answer 3

另一种方法是拥有一个控制通道，任何时候只允许执行一个goroutine，每个例程在完成原子操作时都会发送回“控制锁”：

package main
import "fmt"
import "time"

func routine(id int, control chan struct{}){
    for {
        // Get the control
        <-control
        fmt.Printf("routine %d got control\n", id)
        fmt.Printf("A lot of things happen here...")
        time.Sleep(1)
        fmt.Printf("... but only in routine %d !\n", id)
        fmt.Printf("routine %d gives back control\n", id)
        // Sending back the control to whichever other routine catches it
        control<-struct{}{}
    }
}

func main() {
    // Control channel is blocking
    control := make(chan struct{})

    // Start all routines
    go routine(0, control)
    go routine(1, control)
    go routine(2, control)

    // Sending control to whichever catches it first
    control<-struct{}{}
    // Let routines play for some time...
    time.Sleep(10)
    // Getting control back and terminating
    <-control
    close(control)
    fmt.Println("Finished !")
}

打印：

routine 0 got control
A lot of things happen here...... but only in routine 0 !
routine 0 gives back control
routine 1 got control
A lot of things happen here...... but only in routine 1 !
routine 1 gives back control
routine 2 got control
A lot of things happen here...... but only in routine 2 !
routine 2 gives back control
routine 0 got control
A lot of things happen here...... but only in routine 0 !
routine 0 gives back control
routine 1 got control
A lot of things happen here...... but only in routine 1 !
routine 1 gives back control
routine 2 got control
A lot of things happen here...... but only in routine 2 !
routine 2 gives back control
routine 0 got control
A lot of things happen here...... but only in routine 0 !
routine 0 gives back control
routine 1 got control
A lot of things happen here...... but only in routine 1 !
routine 1 gives back control
routine 2 got control
A lot of things happen here...... but only in routine 2 !
routine 2 gives back control
routine 0 got control
A lot of things happen here...... but only in routine 0 !
routine 0 gives back control
routine 1 got control
A lot of things happen here...... but only in routine 1 !
routine 1 gives back control
routine 2 got control
A lot of things happen here...... but only in routine 2 !
routine 2 gives back control
Finished !

Answer 4

您是否要求显式停止其他例程的库函数？要明确一点，Go和大多数其他语言都不可能。并且同步，互斥情况不适用于您的情况。

并发Goroutines的相互排斥

4 个答案: