Prevent the main() function from terminating before goroutines finish in Golang

Issue

Have loook at this contrived example:

package main

import "fmt"

func printElo() {
    fmt.Printf("Elo\n")
}

func printHello() {
    fmt.Printf("Hello\n")
}

func main() {
    fmt.Printf("This will print.")
    i := 0
    for i < 10 {
        go printElo()
        go printHello()
        i++
    }
}

The output of this program would be just “This will print”. Output of goroutines printElo() and printHello will not be emitted because, I guess, the main() function thread will finish before the goroutines have a chance to even start executing.

What is the idiomatic way to make similar code work in Golang and not terminate prematurely?

Solution

Simplest, cleanest and "scalable" way to do it is to use a sync.WaitGroup:

var wg = &sync.WaitGroup{}

func printElo() {
    defer wg.Done()
    fmt.Printf("Elo\n")
}

func printHello() {
    defer wg.Done()
    fmt.Printf("Hello\n")
}

func main() {
    fmt.Printf("This will print.")
    i := 0
    for i < 10 {
        wg.Add(1)
        go printElo()
        wg.Add(1)
        go printHello()
        i++
    }
    wg.Wait()
}

Output (try it on the Go Playground):

This will print.Hello
Elo
Hello
Elo
Hello
Elo
Hello
Elo
Hello
Elo
Hello
Elo
Hello
Elo
Hello
Elo
Hello
Elo
Hello
Elo

Simple "rules" to follow when doing it with sync.WaitGroup:

  • call WaitGroup.Add() in the "original" goroutine (that starts a new) before the go statement
  • recommended to call WaitGroup.Done() deferred, so it gets called even if the goroutine panics
  • if you want to pass WaitGroup to other functions (and not use a package level variable), you must pass a pointer to it, else the WaitGroup (which is a struct) would be copied, and the Done() method called on the copy wouldn’t be observed on the original

Answered By – icza

Answer Checked By – Mildred Charles (GoLangFix Admin)

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