GMP模型协程调度的练习

package main


import (
    "fmt"
    "runtime"
    "sync"
)

func main() {
    runtime.GOMAXPROCS(1)
    var sync sync.WaitGroup
    sync.Add(6)
    go func() {
        defer sync.Done()
        fmt.Println("goroutine 1")
        go func() {
            defer sync.Done()
            fmt.Println("goroutine 1-1")
            go func() {
                defer sync.Done()
                fmt.Println("goroutine 1-1-1")
            }()
        }()
        go func() {
            defer sync.Done()
            fmt.Println("goroutine 1-2")
            go func() {
                defer sync.Done()
                fmt.Println("goroutine 1-2-1")
            }()
        }()
    }()
    go func() {
        defer sync.Done()
        fmt.Println("goroutine 2")
    }()
    sync.Wait()
    return
}

1. 根据GMP模型，runtime.GOMAXPROCS(1)，P只有一个，也就是只能有一个goroutine运行
2. P的结构中下一个执行的协程runnext和本地队列runq（先进先出）

故结果为：

goroutine 2
goroutine 1
goroutine 1-2
goroutine 1-2-1
goroutine 1-1
goroutine 1-1-1

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