dongxuying7583
dongxuying7583
2016-08-10 13:49
浏览 36
已采纳

golang调度程序如何在以下代码中调度goroutine?

package main

import (
    "fmt"
    "time"
    "runtime"
)

var quit chan int = make(chan int)

func loop(a int) {
    fmt.Println(a)
    for i := 0; i < 30000000000; i++ {
    }
    fmt.Println(a)
    quit <- 0
}

func main() {
    runtime.GOMAXPROCS(1)

    go loop(1)
    time.Sleep(time.Second)
    go loop(2)

    for i := 0; i < 2; i++ {
        <-quit
    }
}

For the scheduler model (M+P+G), I suppose we just have 1 cpu context because we set GOMAXPROCS as 1, and there is just 1 thread(M) here.

In the goroutine, the for loop do not has any IO blocking, so no new thread will be generated, all goroutine should still work in current thread, so I think the 2 goroutine must go one by one, thus, the result should be 1 1 2 2. But in fact, the result is 1 2 1 2. Why?

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1条回答 默认 最新

  • dounao4179
    dounao4179 2016-08-10 14:55
    已采纳

    Here's the order of the relevant operations in your program.

    1. go loop(1) dispatches goroutine loop1
    2. time.Sleep(time.Second) starts in the main goroutine
    3. fmt.Println(a) called from loop1, prints 1
    4. loop1 enters busy loop and holds the CPU
    5. entry to fmt.Println(a) from loop1 yields to the scheduler
    6. main goroutine wakes up, finishes the time.Sleep call
    7. go loop(2) dispatches goroutine loop2
    8. fmt.Println(a) called from loop2, prints 2
    9. loop2 enters busy loop and holds the CPU
    10. entry to fmt.Println(a) from loop2 yields to the scheduler
    11. loop1 one wakes up and finishes fmt.Println(a), prints 1
    12. loop1 sends 0 over the quit channel
    13. loop2 finishes fmt.Println(a), prints 2
    14. loop2 sends 0 over the quit channel
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