Strange behavior of go routine

Question

I just tried the following code, but the result seems a little strange. It prints odd numbers first, and then even numbers. I'm really confused about it. I had hoped it outputs odd number and even number one after another, just like 1, 2, 3, 4... . Who can help me?

package main

import (
    "fmt"
    "time"
)

func main() {
    go sheep(1)
    go sheep(2)
    time.Sleep(100000)
}

func sheep(i int) {
    for ; ; i += 2 {
        fmt.Println(i,"sheeps")
    }
}

Answer 1

More than likely you are only running with one cpu thread. so it runs the first goroutine and then the second. If you tell go it can run on multiple threads then both will be able to run simultaneously provided the os has spare time on a cpu to do so. You can demonstrate this by setting GOMAXPROCS=2 before running your binary. Or you could try adding a runtime.Gosched() call in your sheep function and see if that triggers the runtime to allow the other goroutine to run.

In general though it's better not to assume ordering semantics between operations in two goroutines unless you specify specific synchronization points using a sync.Mutex or communicating between them on channels.

Answer 2

Unsynchronized goroutines execute in a completely undefined order. If you want to print out something like

1 sheeps
2 sheeps
3 sheeps
....

in that exact order, then goroutines are the wrong way to do it. Concurrency works well when you don't care so much about the order in which things occur.

You could impose an order in your program through synchronization (locking a mutex around the fmt.Println calls or using a channel), but it's pointless since you could more easily just write code that uses a single goroutine.