channel面试题。请附上答案,看谁的简单。来玩儿~~~~

xmge · · 1606 次点击
``` package main import ( "fmt" "sync" ) func main() { var ( a = []interface{}{1, 2, 3, 4, 5, 6} b = []interface{}{"a", "b", "c", "d", "e"} wg sync.WaitGroup cha = make(chan struct{}, 1) chb = make(chan struct{}, 1) ) wg.Add(2) go Run(&wg, a, cha, chb) go Run(&wg, b, chb, cha) chb <- struct{}{} wg.Wait() } func Run(group *sync.WaitGroup, data []interface{}, selfChan chan struct{}, peerChan chan struct{}) { for _, d := range data { if _, ok := <-peerChan; !ok { fmt.Println(d) continue } fmt.Println(d) selfChan <- struct{}{} } close(selfChan) group.Done() } ```
#2
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``` func DataIn(a []int64,b []string,ch chan interface{},wg *sync.WaitGroup){ for i:=0;i<4;i++{ ch<-a[i] ch<-b[i] } close(ch) wg.Done() } func DataOut(ch chan interface{},wg *sync.WaitGroup){ for v:=range ch{ fmt.Println(v) } wg.Done() } func main(){ var wg sync.WaitGroup arra := []int64{1,2,3,4} arrb := []string{"a","b","c","d"} ch:=make(chan interface{}) wg.Add(2) go DataIn(arra,arrb,ch,&wg) go DataOut(ch,&wg) wg.Wait() } ```
#1
//我的最简单吧 ``` package main import ( "fmt" "time" ) func PrintA(arra []int64,ch chan int,lock chan bool){ for i,v:=range arra{ fmt.Println(v) ch<-i lock<-true } } func PrintB(arrb []string,ch chan int,lock chan bool) { for { fmt.Println(arrb[<-ch]) <-lock } } func main(){ arra := []int64{1,2,3,4} arrb := []string{"a","b","c","d"} ch:=make(chan int) lock:=make(chan bool) go PrintA(arra,ch,lock) go PrintB(arrb,ch,lock) t:=time.NewTicker(time.Second) for { select{ case <-t.C: fmt.Println("over") return } } } ```
#3