duanche4578
2014-08-01 06:03
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Golang:Goroutine无限循环

When an fmt.Print() line is removed from the code below, code runs infinitely. Why?

package main

import "fmt"
import "time"
import "sync/atomic"

func main() {
        var ops uint64 = 0 
        for i := 0; i < 50; i++ {
                go func() {
                        for {
                                atomic.AddUint64(&ops, 1)
                                fmt.Print()
                        }
                }()
        }
        time.Sleep(time.Second)
        opsFinal := atomic.LoadUint64(&ops)
        fmt.Println("ops:", opsFinal)
}
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1条回答 默认 最新

  • dqkf49487 2014-08-01 06:13
    已采纳

    The Go By Example article includes:

       // Allow other goroutines to proceed.
       runtime.Gosched()
    

    The fmt.Print() plays a similar role, and allows the main() to have a chance to proceed.

    A export GOMAXPROCS=2 might help the program to finish even in the case of an infinite loop, as explained in "golang: goroute with select doesn't stop unless I added a fmt.Print()".

    fmt.Print() explicitly passes control to some syscall stuff


    Yes, go1.2+ has pre-emption in the scheduler

    In prior releases, a goroutine that was looping forever could starve out other goroutines on the same thread, a serious problem when GOMAXPROCS provided only one user thread.

    In Go 1.2, this is partially addressed: The scheduler is invoked occasionally upon entry to a function. This means that any loop that includes a (non-inlined) function call can be pre-empted, allowing other goroutines to run on the same thread.

    Notice the emphasis (that I put): it is possible that in your example the for loop atomic.AddUint64(&ops, 1) is inlined. No pre-emption there.


    Update 2017: Go 1.10 will get rid of GOMAXPROCS.

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