var wg sync.WaitGroup
wg.Add(len(work))
sem := make(chan struct{}, 10)
wgDone := make(chan bool)
for i < len(work)-1 {
go func() {
defer wg.Done()
sem <- struct{}{}
defer func() {
<-sem
}()
worker(work[i])
}()
i = i + 1
}
go func() {
wg.Wait()
close(wgDone)
}()
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我一次只需要10 个新的 goroutine来执行这项工作。这是我当前的解决方案,它阻止 goroutine 继续运行,因此一次只有 10 个。我怎样才能改变这个,这样它就不会创建大量被阻塞等待工作的 goroutine,而是只创建 10 个完成所有工作的 goroutine?
was*_*mup 12
根据用例,以下方法之一很有用:
max新 goroutine 的数量和一个通道作为队列(Go 游乐场):package main
import (
"fmt"
"sync"
)
func main() {
const max = 10
queue := make(chan int, max)
wg := &sync.WaitGroup{}
for i := 0; i < max; i++ {
wg.Add(1)
go worker(wg, queue)
}
for i := 0; i < 100; i++ {
queue <- i
}
close(queue)
wg.Wait()
fmt.Println("Done")
}
func worker(wg *sync.WaitGroup, queue chan int) {
defer wg.Done()
for job := range queue {
fmt.Print(job, " ") // a job
}
}
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package main
import (
"fmt"
"sync"
)
func main() {
const max = 10
semaphore := make(chan struct{}, max)
wg := &sync.WaitGroup{}
for i := 0; i < 1000; i++ {
semaphore <- struct{}{} // acquire
wg.Add(1)
go limited(i, wg, semaphore)
}
wg.Wait()
fmt.Println("Done")
}
func limited(i int, wg *sync.WaitGroup, semaphore chan struct{}) {
defer wg.Done()
fmt.Println("i =", i) // a job
<-semaphore // release
}
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package main
import (
"fmt"
"sync"
)
func main() {
const max = 10
semaphore := make(chan struct{}, max)
wg := &sync.WaitGroup{}
for i := 0; i < 1000; i++ {
wg.Add(1)
go limited(i, wg, semaphore)
}
wg.Wait()
fmt.Println("Done")
}
func limited(i int, wg *sync.WaitGroup, semaphore chan struct{}) {
defer wg.Done()
semaphore <- struct{}{} // acquire
fmt.Println("i =", i) // a job
<-semaphore // release
}
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