两个goroutine 循环打印1~10

jxy_90 · 2022-03-07 20:33:33 · 7740 次点击

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
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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

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