How to collect values from a channel into a slice in Go? - go

Question

Suppose I have a helper function helper(n int) which returns a slice of integers of variable length. I would like to run helper(n) in parallel for various values of n and collect the output in one big slice. My first attempt at this is the following:
package main
import (
"fmt"
"golang.org/x/sync/errgroup"
)
func main() {
out := make([]int, 0)
ch := make(chan int)
go func() {
for i := range ch {
out = append(out, i)
}
}()
g := new(errgroup.Group)
for n := 2; n <= 3; n++ {
n := n
g.Go(func() error {
for _, i := range helper(n) {
ch <- i
}
return nil
})
}
if err := g.Wait(); err != nil {
panic(err)
}
close(ch)
// time.Sleep(time.Second)
fmt.Println(out) // should have the same elements as [0 1 0 1 2]
}
func helper(n int) []int {
out := make([]int, 0)
for i := 0; i < n; i++ {
out = append(out, i)
}
return out
}
However, if I run this example I do not get all 5 expected values, instead I get
[0 1 0 1]
(If I uncomment the time.Sleep I do get all five values, [0 1 2 0 1], but this is not an acceptable solution).
It seems that the problem with this is that out is being updated in a goroutine, but the main function returns before it is done updating.
One thing that would work is using a buffered channel of size 5:
func main() {
ch := make(chan int, 5)
g := new(errgroup.Group)
for n := 2; n <= 3; n++ {
n := n
g.Go(func() error {
for _, i := range helper(n) {
ch <- i
}
return nil
})
}
if err := g.Wait(); err != nil {
panic(err)
}
close(ch)
out := make([]int, 0)
for i := range ch {
out = append(out, i)
}
fmt.Println(out) // should have the same elements as [0 1 0 1 2]
}
However, although in this simplified example I know what the size of the output should be, in my actual application this is not known a priori. Essentially what I would like is an 'infinite' buffer such that sending to the channel never blocks, or a more idiomatic way to achieve the same thing; I've read https://blog.golang.org/pipelines but wasn't able to find a close match to my use case. Any ideas?

Answer 1

In this version of the code, the execution is blocked until ch is closed.
ch is always closed at the end of a routine that is responsible to push into ch. Because the program pushes to ch in a routine, it is not needed to use a buffered channel.
package main
import (
"fmt"
"golang.org/x/sync/errgroup"
)
func main() {
ch := make(chan int)
go func() {
g := new(errgroup.Group)
for n := 2; n <= 3; n++ {
n := n
g.Go(func() error {
for _, i := range helper(n) {
ch <- i
}
return nil
})
}
if err := g.Wait(); err != nil {
panic(err)
}
close(ch)
}()
out := make([]int, 0)
for i := range ch {
out = append(out, i)
}
fmt.Println(out) // should have the same elements as [0 1 0 1 2]
}
func helper(n int) []int {
out := make([]int, 0)
for i := 0; i < n; i++ {
out = append(out, i)
}
return out
}
Here is the fixed version of the first code, it is convoluted but demonstrates the usage of sync.WaitGroup.
package main
import (
"fmt"
"sync"
"golang.org/x/sync/errgroup"
)
func main() {
out := make([]int, 0)
ch := make(chan int)
var wg sync.WaitGroup
wg.Add(1)
go func() {
defer wg.Done()
for i := range ch {
out = append(out, i)
}
}()
g := new(errgroup.Group)
for n := 2; n <= 3; n++ {
n := n
g.Go(func() error {
for _, i := range helper(n) {
ch <- i
}
return nil
})
}
if err := g.Wait(); err != nil {
panic(err)
}
close(ch)
wg.Wait()
// time.Sleep(time.Second)
fmt.Println(out) // should have the same elements as [0 1 0 1 2]
}
func helper(n int) []int {
out := make([]int, 0)
for i := 0; i < n; i++ {
out = append(out, i)
}
return out
}