英文:
Sync between 2 goroutines
问题
我的任务是同步两个goroutine,使输出看起来像这样:
foobarfoobarfoobarfoobar
问题在于当我调用它们时,它们的输出完全是随机的。这是我的代码:
package main
import (
"fmt"
"sync"
"time"
)
type ConcurrentPrinter struct {
sync.WaitGroup
sync.Mutex
}
func (cp *ConcurrentPrinter) printFoo(times int) {
cp.WaitGroup.Add(times)
go func() {
cp.Lock()
fmt.Print("foo")
cp.Unlock()
}()
}
func (cp *ConcurrentPrinter) printBar(times int) {
cp.WaitGroup.Add(times)
go func() {
cp.Lock()
fmt.Print("bar")
cp.Unlock()
}()
}
func main() {
times := 10
cp := &ConcurrentPrinter{}
for i := 0; i <= times; i++ {
cp.printFoo(i)
cp.printBar(i)
}
time.Sleep(10 * time.Millisecond)
}
英文:
My task is to sync 2 goroutines so the output should look like that:
> foobarfoobarfoobarfoobar
.The issue is that when I call them they come out completely randomized. This is my code:
package main
import (
"fmt"
"sync"
"time"
)
type ConcurrentPrinter struct {
sync.WaitGroup
sync.Mutex
}
func (cp *ConcurrentPrinter) printFoo(times int) {
cp.WaitGroup.Add(times)
go func() {
cp.Lock()
fmt.Print("foo")
cp.Unlock()
}()
}
func (cp *ConcurrentPrinter) printBar(times int) {
cp.WaitGroup.Add(times)
go func() {
cp.Lock()
fmt.Print("bar")
cp.Unlock()
}()
}
func main() {
times := 10
cp := &ConcurrentPrinter{}
for i := 0; i <= times; i++ {
cp.printFoo(i)
cp.printBar(i)
}
time.Sleep(10 * time.Millisecond)
}
答案1
得分: 2
如评论中所述,使用goroutine可能不是你试图实现的最佳方案,因此这可能是一个XY问题。
话虽如此,如果你想确保两个独立的goroutine以交替的顺序进行工作,你可以实现一组“乒乓”互斥锁:
var ping, pong sync.Mutex
pong.Lock() // 确保第二个goroutine等待,第一个先执行
go func() {
for {
ping.Lock()
foo()
pong.Unlock()
}
}()
go func() {
for {
pong.Lock()
bar()
ping.Unlock()
}
}()
https://go.dev/play/p/VO2LoMJ8fek
英文:
As outlined in the comments, using goroutines may not be the best use case for what you are trying to achieve - and thus this may be an XY problem.
Having said that, if you want to ensure two independent goroutines interleave their work in an alternating sequence, you can implement a set of "ping-pong" mutexs:
var ping, pong sync.Mutex
pong.Lock() // ensure the 2nd goroutine waits & the 1st goes first
go func() {
for {
ping.Lock()
foo()
pong.Unlock()
}
}()
go func() {
for {
pong.Lock()
bar()
ping.Unlock()
}
}()
答案2
得分: 0
使用通道:
func printFoo(i int, ch chan<- bool, wg *sync.WaitGroup) {
wg.Add(1)
go func() {
defer wg.Done()
fmt.Print("foo")
ch <- true
}()
}
func printBar(i int, ch chan<- bool, wg *sync.WaitGroup) {
wg.Add(1)
go func() {
defer wg.Done()
fmt.Print("bar")
ch <- true
}()
}
func main() {
times := 4
firstchan := make(chan bool)
secondchan := make(chan bool)
var wg sync.WaitGroup
for i := 0; i <= times; i++ {
printFoo(i, firstchan, &wg)
<-firstchan
printBar(i, secondchan, &wg)
<-secondchan
}
wg.Wait()
}
https://go.dev/play/p/MlZ9dHkUXGb
英文:
Using channel:
func printFoo(i int, ch chan<- bool, wg *sync.WaitGroup) {
wg.Add(1)
go func() {
defer wg.Done()
fmt.Print("foo")
ch <- true
}()
}
func printBar(i int, ch chan<- bool, wg *sync.WaitGroup) {
wg.Add(1)
go func() {
defer wg.Done()
fmt.Print("bar")
ch <- true
}()
}
func main() {
times := 4
firstchan := make(chan bool)
secondchan := make(chan bool)
var wg sync.WaitGroup
for i := 0; i <= times; i++ {
printFoo(i, firstchan, &wg)
<-firstchan
printBar(i, secondchan, &wg)
<-secondchan
}
wg.Wait()
}
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