What is the method set of `sync.WaitGroup`?

Question

I have this simple program below

package main

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

var wg sync.WaitGroup

func main() {
    wg.Add(1)

    go func() {
        fmt.Println("starting...")
        time.Sleep(1 * time.Second)
        fmt.Println("done....")
        wg.Done()
    } ()

    wg.Wait()

}

Notice that I use var wg sync.WaitGroup as a value, not a pointer. But the page for the sync package specifies that the Add, Done and Wait function take a *sync.WaitGroup.

Why/How does this work?

Answer 1

The method set of sync.WaitGroup is the empty method set:

wg := sync.WaitGroup{}
fmt.Println(reflect.TypeOf(wg).NumMethod())

Output (try it on the Go Playground):

0

This is because all the methods of sync.WaitGroup have pointer receivers, so they are all part of the method set of the *sync.WaitGroup type.

When you do:

var wg sync.WaitGroup

wg.Add(1)
wg.Done()
// etc.

This is actually a shorthand for (&wg).Add(1), (&wg).Done() etc.

This is in Spec: Calls:

If x is addressable and &x's method set contains m, x.m() is shorthand for (&x).m().

So when you have a value that is addressable (a variable is addressable), you may call any methods that have pointer receiver on non-pointer values, and the compiler will automatically take the address and use that as the receiver value.

See related question:

Calling a method with a pointer receiver by an object instead of a pointer to it?

Answer 2

In your case you are modifying a global wg object, if you pass it to a function you have to use pointer because you need to modify the object itself. If you pass by value, inside your function you will be modifying a copy of it, not the object itself.