戈朗 : goroutine infinite-loop

Question

当从下面的代码中删除 fmt.Print() 行时，代码将无限运行。为什么？

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)
}

Answer 1

Go By Example article includes :

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

fmt.Print() 起着类似的作用，并允许 main() 有机会继续执行。

即使在无限循环的情况下，export GOMAXPROCS=2 也可能有助于程序完成，如“golang: goroute with select doesn't stop unless I added a fmt.Print()”中所述。

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

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是的，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.

请注意(我强调的)重点:在您的示例中，for 循环 atomic.AddUint64(&ops, 1) 可能是内联的。那里没有先发制人。

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2017 年更新:Go 1.10 will get rid of GOMAXPROCS .