我正在開發(fā)一個(gè)小型實(shí)用程序，它需要迭代動(dòng)態(tài)范圍的項(xiàng)目（可以是 100 或可以是 100000）并將這些項(xiàng)目放入通道中。另一個(gè)函數(shù)從該通道讀取項(xiàng)目并單獨(dú)對(duì)每個(gè)項(xiàng)目進(jìn)行一些處理。我試圖用來sync.WaitGroup確保在處理通道中的所有項(xiàng)目之前我的實(shí)用程序不會(huì)退出。由于我對(duì)頻道和等待組相當(dāng)陌生，因此我遇到了錯(cuò)誤panic: sync: WaitGroup is reused before previous Wait has returnedhttps://play.golang.org/p/nMw3END_9qwpackage mainimport ( "fmt" "github.com/dchest/uniuri" "math/rand" "sync" "time")var wg sync.WaitGroupvar count = 0func printMe(msg string) { time.Sleep(1 * time.Second) fmt.Println(count, msg)}func makeMePrint(ch chan string) { for s := range ch { count++ wg.Add(1) printMe(s) wg.Done() }}func writePrint(ch chan<- string) { fmt.Println("Starting to insert data in channel") for i := 1; i <= rand.Intn(30); i++ { s := uniuri.New() ch <- s } fmt.Println("We are done inserting all data in the channel") close(ch)}func main() { var ch = make(chan string) go writePrint(ch) go makeMePrint(ch) time.Sleep(1 * time.Second) wg.Wait()}這是我正在研究的主要思想（不是確切的代碼，而是具有相同數(shù)量功能的完全相同的架構(gòu)）。如何確保該實(shí)用程序僅在通道中的所有項(xiàng)目都是進(jìn)程時(shí)才退出。任何幫助表示贊賞。
1 回答
富國(guó)滬深
TA貢獻(xiàn)1790條經(jīng)驗(yàn) 獲得超9個(gè)贊
我終于讓它發(fā)揮作用了。
package main
import (
"fmt"
"github.com/dchest/uniuri"
"math/rand"
"sync"
"time"
)
var count = 0
func printMe(msg string) {
time.Sleep(1 * time.Second)
fmt.Println(count, msg)
}
func makeMePrint(wg *sync.WaitGroup, ch chan string) {
wg.Add(1)
defer wg.Done()
for s := range ch {
count++
printMe(s)
}
}
func writePrint(wg *sync.WaitGroup, ch chan<- string) {
wg.Add(1)
defer wg.Done()
fmt.Println("Starting to insert data in channel")
for i := 1; i <= rand.Intn(30); i++ {
s := uniuri.New()
ch <- s
}
fmt.Println("We are done inserting all data in the channel")
close(ch)
}
func main() {
wg := &sync.WaitGroup{}
var ch = make(chan string)
go writePrint(wg, ch)
go makeMePrint(wg, ch)
time.Sleep(1 * time.Second)
wg.Wait()
}
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