How to wait for all goroutines to finish without u

Question

This code selects all xml files in the same folder, as the invoked executable and asynchronously applies processing to each result in the callback method (in the example below, just the name of the file is printed out).

How do I avoid using the sleep method to keep the main method from exiting? I have problems wrapping my head around channels (I assume that's what it takes, to synchronize the results) so any help is appreciated!

package main

import (
    "fmt"
    "io/ioutil"
    "path"
    "path/filepath"
    "os"
    "runtime"
    "time"
)

func eachFile(extension string, callback func(file string)) {
    exeDir := filepath.Dir(os.Args[0])
    files, _ := ioutil.ReadDir(exeDir)
    for _, f := range files {
            fileName := f.Name()
            if extension == path.Ext(fileName) {
                go callback(fileName)
            }
    }
}


func main() {
    maxProcs := runtime.NumCPU()
    runtime.GOMAXPROCS(maxProcs)

    eachFile(".xml", func(fileName string) {
                // Custom logic goes in here
                fmt.Println(fileName)
            })

    // This is what i want to get rid of
    time.Sleep(100 * time.Millisecond)
}

Answer 1

WaitGroups are definitely the canonical way to do this. Just for the sake of completeness, though, here's the solution that was commonly used before WaitGroups were introduced. The basic idea is to use a channel to say "I'm done," and have the main goroutine wait until each spawned routine has reported its completion.

func main() {
    c := make(chan struct{}) // We don't need any data to be passed, so use an empty struct
    for i := 0; i < 100; i++ {
        go func() {
            doSomething()
            c <- struct{}{} // signal that the routine has completed
        }()
    }

    // Since we spawned 100 routines, receive 100 messages.
    for i := 0; i < 100; i++ {
        <- c
    }
}

Answer 2

sync.WaitGroup can help you here.

package main

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


func wait(seconds int, wg * sync.WaitGroup) {
    defer wg.Done()

    time.Sleep(time.Duration(seconds) * time.Second)
    fmt.Println("Slept ", seconds, " seconds ..")
}


func main() {
    var wg sync.WaitGroup

    for i := 0; i <= 5; i++ {
        wg.Add(1)   
        go wait(i, &wg)
    }
    wg.Wait()
}

Answer 3

You can use sync.WaitGroup. Quoting the linked example:

package main

import (
        "net/http"
        "sync"
)

func main() {
        var wg sync.WaitGroup
        var urls = []string{
                "http://www.golang.org/",
                "http://www.google.com/",
                "http://www.somestupidname.com/",
        }
        for _, url := range urls {
                // Increment the WaitGroup counter.
                wg.Add(1)
                // Launch a goroutine to fetch the URL.
                go func(url string) {
                        // Decrement the counter when the goroutine completes.
                        defer wg.Done()
                        // Fetch the URL.
                        http.Get(url)
                }(url)
        }
        // Wait for all HTTP fetches to complete.
        wg.Wait()
}