在我的代碼中，有三個并發(fā)例程。我嘗試簡要概述一下我的代碼，Routine 1 {do something*Send int to Routine 2Send int to Routine 3Print SomethingPrint Something*do something}Routine 2 {do something*Send int to Routine 1Send int to Routine 3Print SomethingPrint Something*do something}Routine 3 {do something*Send int to Routine 1Send int to Routine 2Print SomethingPrint Something*do something}main {routine1routine2routine3}我希望在執(zhí)行兩個代碼之間的某些操作（兩個星號之間的代碼）時，控制流一定不能進(jìn)入其他goroutine。例如，當(dāng)例程1正在執(zhí)行兩顆星之間的事件（發(fā)送和打印事件）時，例程2和3必須被阻止（執(zhí)行的平均流量不會從例程1傳遞到例程2或3）。在完成上一個打印事件后，執(zhí)行流程可能會傳遞到例程2或3。有人可以通過指定幫助我的方式，我該如何實(shí)現(xiàn)呢？是否可以通過WaitGroup來實(shí)現(xiàn)上述規(guī)范？任何人都可以通過舉一個簡單的示例向我展示如何使用WaitGroup來實(shí)現(xiàn)上面指定的示例。謝謝。注意：我提供兩個發(fā)送和兩個打印選項(xiàng)，實(shí)際上有很多發(fā)送和打印功能。
3 回答
瀟湘沐
TA貢獻(xiàn)1816條經(jīng)驗(yàn) 獲得超6個贊
如果我正確理解了它，那么您想要的是防止同時執(zhí)行每個功能的某些部分和其他功能。下面的代碼可以做到這一點(diǎn)：fmt.Println當(dāng)其他例程正在運(yùn)行時，行不會發(fā)生。這是發(fā)生的情況：當(dāng)執(zhí)行到達(dá)打印部分時，它將等待直到其他例程結(jié)束（如果它們正在運(yùn)行），并且在執(zhí)行此打印行時，其他例程不會啟動并等待。我希望這就是您想要的。如果我對此有誤，請糾正我。
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：讓我在這里解釋這三個互斥鎖：一個互斥鎖，如文檔所述：“互斥鎖”。這意味著當(dāng)您調(diào)用Lock互斥鎖時，如果其他人已鎖定相同的互斥鎖，則代碼僅在此處等待。在您致電之后Unlock，被阻止的代碼將立即恢復(fù)。
在這里，我通過在函數(shù)的開頭鎖定互斥鎖并將其解鎖來將每個函數(shù)置于其自己的互斥鎖中。通過這種簡單的機(jī)制，您可以避免與這些功能同時運(yùn)行所需的任何代碼部分。例如，在您想要擁有運(yùn)行時不應(yīng)該運(yùn)行的代碼的任何地方Routine1，只需mutex1在該代碼的開頭鎖定并在最后解鎖即可。這就是我在Routine2和中的適當(dāng)行中所做的事情Routine3。希望能澄清一些事情。
一只甜甜圈
TA貢獻(xiàn)1836條經(jīng)驗(yàn) 獲得超5個贊
另一種方法是擁有一個控制通道，在任何時候只允許執(zhí)行一個goroutine，并且每當(dāng)它們完成其原子操作時，每個例程就會發(fā)回“ control lock”：
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?。?/p>
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 !
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