等待n个goroutine终止

Question

I need to start a huge amount of goroutines and wait for their termination. The intuitive way seems to use a channel to wait till all of them are finished :

package main

type Object struct {
    //data
}

func (obj *Object) Update(channel chan int) {
    //update data
    channel <- 1
    return
}

func main() {

    channel := make(chan int, n)
    list := make([]Object, n, m)
    for {
        for _, object := range list {
            go object.Update(channel)
        }
        for i := 0; i < n; i++ {
            <-channel
        }
        //now everything has been updated. start again
    }
}

But the problem is that the amount of objects and therefore the amount of goroutines could change. Is it possible to change the buffer size of a channel ?

Is there maybe a more elegant way to do this ?

Answer 1

I've used WaitGroup as a solution to this problem. Translating your current code, with some logs to make it clear what's happening:

package main

import "sync"
import "fmt"
import "time"

type Object struct {
    //data
}

func (obj *Object) Update(wg *sync.WaitGroup) {
    //update data
    time.Sleep(time.Second)
    fmt.Println("Update done")
    wg.Done()
    return
}

func main() {
    var wg sync.WaitGroup
    list := make([]Object, 5)
    for {
        for _, object := range list {
            wg.Add(1)
            go object.Update(&wg)
        }
        //now everything has been updated. start again
        wg.Wait()
        fmt.Println("Group done")
    }
}