sync.Waitgroup不会阻止执行

Question

Consider this code snippet

package main

import (
    "fmt"
    "sync"
    "time"
)

func main() {
    wg := new(sync.WaitGroup)
    nap := func() {
        wg.Add(1)
        time.Sleep(2 * time.Second)
        fmt.Println("nap done")
        wg.Done()
    }

    go nap()
    go nap()
    go nap()

    fmt.Println("nap time")
    wg.Wait()
    fmt.Println("all done")
}

Running such code gives expected output:

nap time
nap done
nap done
nap done
all done

Now let's ommit first standard output print before wg.Wait():

// fmt.Println("nap time")
wg.Wait()
fmt.Println("all done")

The output now changes to unexpected:

all done

Where expected would be:

nap done
nap done
nap done
all done

Same code on the playground does give this output without a need of omitting the stdout print.

Can you explain to me, what I am missing there?

Answer 1

Even though this looks like magic, it has a logical explanation. Go doesn't guarantee the order of goroutines execution. There are three goroutines spawned in given snippet code, but in fact, there are four of them: the very first one that is spawned when the execution starts.

Stdout print is omitted

This goroutine spawned three nap functions and continued in its plan. It was so fast that it executed wg.Wait() before any of spawned goroutines was able to call wg.Add(1). As a result wg.Wait() didn't block the execution and program ended.

Print to stdout before wg.Wait()

In this case, program execution was different, goroutines were able to make wg.Add(1) call because the main goroutine wasn't fast as in the first case. This behavior isn't guaranteed, which can be seen in the linked playground example.

It has nothing to do with stdout print

The following code sample will give the same expected output:

time.Sleep(time.Second)
wg.Wait()
fmt.Println("all done")

The fmt.Println() had the same impact as time.Sleep() had.

Idiomatic way

The rule is simple: call wg.Add(1) before spawning the goroutine.

package main

import (
    "fmt"
    "sync"
    "time"
)

func main() {
    wg := new(sync.WaitGroup)
    nap := func() {
        time.Sleep(2 * time.Second)
        fmt.Println("nap done")
        wg.Done()
    }

    napCount := 3
    wg.Add(napCount)
    for i := 0; i < napCount; i++ {
        go nap()
    }

    wg.Wait()
    fmt.Println("all done")
}