[英]In Golang, how to handle many goroutines with channel
我想在 Golang 中使用 for 循環同時啟動 1000 個 goroutine。
問題是:我必須確保每個 goroutine 都已執行。
是否可以使用渠道來幫助我確定這一點?
結構有點像這樣:
func main {
for i ... {
go ...
ch?
ch?
}
正如@Andy 提到的,您可以使用sync.WaitGroup
來實現這一點。 下面是一個例子。 希望代碼是不言自明的。
package main
import (
"fmt"
"sync"
"time"
)
func dosomething(millisecs int64, wg *sync.WaitGroup) {
defer wg.Done()
duration := time.Duration(millisecs) * time.Millisecond
time.Sleep(duration)
fmt.Println("Function in background, duration:", duration)
}
func main() {
arr := []int64{200, 400, 150, 600}
var wg sync.WaitGroup
for _, n := range arr {
wg.Add(1)
go dosomething(n, &wg)
}
wg.Wait()
fmt.Println("Done")
}
我建議你遵循一個模式。 Concurrency 和 Channel 是好的,但是如果您以不好的方式使用它,您的程序可能會變得比預期的更慢。 處理多個 go-routine 和 channel 的簡單方法是通過工作池模式。
仔細看看下面的代碼
// In this example we'll look at how to implement
// a _worker pool_ using goroutines and channels.
package main
import "fmt"
import "time"
// Here's the worker, of which we'll run several
// concurrent instances. These workers will receive
// work on the `jobs` channel and send the corresponding
// results on `results`. We'll sleep a second per job to
// simulate an expensive task.
func worker(id int, jobs <-chan int, results chan<- int) {
for j := range jobs {
fmt.Println("worker", id, "started job", j)
time.Sleep(time.Second)
fmt.Println("worker", id, "finished job", j)
results <- j * 2
}
}
func main() {
// In order to use our pool of workers we need to send
// them work and collect their results. We make 2
// channels for this.
jobs := make(chan int, 100)
results := make(chan int, 100)
// This starts up 3 workers, initially blocked
// because there are no jobs yet.
for w := 1; w <= 3; w++ {
go worker(w, jobs, results)
}
// Here we send 5 `jobs` and then `close` that
// channel to indicate that's all the work we have.
for j := 1; j <= 5; j++ {
jobs <- j
}
close(jobs)
// Finally we collect all the results of the work.
for a := 1; a <= 5; a++ {
<-results
}
}
這個簡單的例子取自這里。 此外, results
通道可以幫助您跟蹤執行作業的所有 go 例程,包括失敗通知。
要確保 goroutine 完成並收集結果,請嘗試以下示例:
package main
import (
"fmt"
)
const max = 1000
func main() {
for i := 1; i <= max; i++ {
go f(i)
}
sum := 0
for i := 1; i <= max; i++ {
sum += <-ch
}
fmt.Println(sum) // 500500
}
func f(n int) {
// do some job here and return the result:
ch <- n
}
var ch = make(chan int, max)
為了等待 1000 個 goroutines 完成,試試這個例子:
package main
import (
"fmt"
"sync"
)
func main() {
wg := &sync.WaitGroup{}
for i := 0; i < 1000; i++ {
wg.Add(1)
go f(wg, i)
}
wg.Wait()
fmt.Println("Done.")
}
func f(wg *sync.WaitGroup, n int) {
defer wg.Done()
fmt.Print(n, " ")
}
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