Example for sync.WaitGroup correct?

Is this example usage of sync.WaitGroup correct? It gives the expected result, but I am unsure about the wg.Add(4) and the position of wg.Done(). Does it make sense to add the four goroutines at once with wg.Add()?

http://play.golang.org/p/ecvYHiie0P

package main

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

func dosomething(millisecs time.Duration, wg *sync.WaitGroup) {
	duration := millisecs * time.Millisecond
	time.Sleep(duration)
	fmt.Println("Function in background, duration:", duration)
	wg.Done()
}

func main() {
	var wg sync.WaitGroup
	wg.Add(4)
	go dosomething(200, &wg)
	go dosomething(400, &wg)
	go dosomething(150, &wg)
	go dosomething(600, &wg)

	wg.Wait()
	fmt.Println("Done")
}

Result (as expected):

Function in background, duration: 150ms
Function in background, duration: 200ms
Function in background, duration: 400ms
Function in background, duration: 600ms
Done

Yes, this example is correct. It is important that the wg.Add() happens before the go statement to prevent race conditions. The following would also be correct:

func main() {
    var wg sync.WaitGroup
    wg.Add(1)
    go dosomething(200, &wg)
    wg.Add(1)
    go dosomething(400, &wg)
    wg.Add(1)
    go dosomething(150, &wg)
    wg.Add(1)
    go dosomething(600, &wg)

    wg.Wait()
    fmt.Println("Done")
}

However, it is rather pointless to call wg.Add over and over again when you already know how many times it will be called.

Waitgroups panic if the counter falls below zero. The counter starts at zero, each Done() is a -1 and each Add() depends on the parameter. So, to ensure that the counter never drops below and avoid panics, you need the Add() to be guaranteed to come before the Done().

In Go, such guarantees are given by the memory model.

The memory model states that all statements in a single goroutine appear to be executed in the same order as they are written. It is possible that they won't actually be in that order, but the outcome will be as if it was. It is also guaranteed that a goroutine doesn't run until after the go statement that calls it. Since the Add() occurs before the go statement and the go statement occurs before the Done(), we know the Add() occurs before the Done().

If you were to have the go statement come before the Add(), the program may operate correctly. However, it would be a race condition because it would not be guaranteed.

I would recommend embeding the wg.Add() call into the doSomething() function itself, so that if you adjust the number of times it's called, you don't have to separately adjust the add parameter manually which could lead to an error if you update one but forget to update the other (in this trivial example that is unlikely, but still, I personally believe it to be better practice for code re-use).

As Stephen Weinberg points out in his answer to this question, you do have to increment the waitgroup prior to spawning the gofunc, but you can accomplish this easily by wrapping the gofunc spawn inside the doSomething() function itself, like this:

func dosomething(millisecs time.Duration, wg *sync.WaitGroup) {
    wg.Add(1)
    go func() {
	    duration := millisecs * time.Millisecond
	    time.Sleep(duration)
	    fmt.Println("Function in background, duration:", duration)
	    wg.Done()
    }()
}

Then you can call it without the go invocation, e.g.:

func main() {
    var wg sync.WaitGroup
    dosomething(200, &wg)
    dosomething(400, &wg)
    dosomething(150, &wg)
    dosomething(600, &wg)
    wg.Wait()
    fmt.Println("Done")
}

As a playground: http://play.golang.org/p/WZcprjpHa_

• small improvement based on Mroth answer
• using defer for Done is safer

> func dosomething(millisecs time.Duration, wg *sync.WaitGroup) {
> wg.Add(1)
> go func() {
> defer wg.Done()
> duration := millisecs * time.Millisecond
> time.Sleep(duration)
> fmt.Println("Function in background, duration:", duration)
> }()
> }
>
> func main() {
> var wg sync.WaitGroup
> dosomething(200, &wg)
> dosomething(400, &wg)
> dosomething(150, &wg)
> dosomething(600, &wg)
> wg.Wait()
> fmt.Println("Done")
> }