两个goroutine 循环打印1~10

jxy_90 · · 7431 次点击
```go func Test_pingpang(t *testing.T) { const N = 100 ping := make(chan int, 1) pang := make(chan int, 1) done := make(chan [0]int) pingpang := func(in <-chan int, out chan<- int) { // 根本不需要关心到底是+=1还是+=2 for x := range in { t.Log(x) if x < N { out <- (x + 1) continue } close(done) break } } ping <- 0 // 发球 // 也不需要关心多少goroutine,甚至ping/pang也可以不配对。以下是斗地主3打1 go pingpang(ping, pang) go pingpang(pang, ping) go pingpang(ping, pang) go pingpang(ping, pang) <-done close(ping) close(pang) } ```
#17
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现在定义上面的协程为 g1 现在定义下面的协程为 g2 你这段逻辑,不管怎么输出,其执行的顺序一定是 g1 g2 g2 g1 g1 g2 g2 g1 ... 按照这个线性逻辑,去和你的输出做匹配,你就会发现问题所在
#1
```golang package main import ( "fmt" "sync" ) func main() { //fmt.Println("123") c1 := make(chan int) wg := sync.WaitGroup{} wg.Add(2) go func() { for i := 0; i < 10; i += 2 { fmt.Println("go1 sending ") c1 <- 1 fmt.Println("go1 sent # test") } wg.Done() }() go func() { for i := 0; i < 10; i += 2 { fmt.Println("go2 receiving ") <-c1 fmt.Println("go2 received # test") } wg.Done() }() wg.Wait() } ``` 结果如下 带 # test 就是你打印的那部分 ```bash go2 receiving go1 sending go1 sent # test go1 sending go2 received # test go2 receiving go2 received # test go2 receiving go1 sent # test go1 sending go1 sent # test go1 sending go2 received # test go2 receiving go2 received # test go2 receiving go1 sent # test go1 sending go1 sent # test go2 received # test ``` 可以看到,一旦完成一次 channel 的通信后,如果当前 goroutine 还在占用 processor, 那么就会继续运行,直到下一次阻塞
#2