Golang: goroutine infinite-loop

Question

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)
}

VonC · Accepted Answer · 2017-09-28 11:36:20Z

8

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.

edited Sep 28, 2017 at 11:36

answered Aug 1, 2014 at 6:13

VonC

1.4m569 gold badges4.8k silver badges5.7k bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

xged Over a year ago

This from golang.org: 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.

VonC Over a year ago

@xged it wouldn't work in playground though, as commented in stackoverflow.com/a/20836784/6309

VonC Over a year ago

@xged I have edited the answer: if the for loop is inlined, you won't have any pre-emption.

VonC Over a year ago

@xged probably because GOMAXPROCS is always 1 and, since it doesn't work outside playground (because of inlining), it wouldn't work inside playground.

Collectives™ on Stack Overflow

Golang: goroutine infinite-loop

1 Answer 1

4 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

4 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related