Parallel processing in golang - parallel-processing

Question

Given the following code:
package main
import (
"fmt"
"math/rand"
"time"
)
func main() {
for i := 0; i < 3; i++ {
go f(i)
}
// prevent main from exiting immediately
var input string
fmt.Scanln(&input)
}
func f(n int) {
for i := 0; i < 10; i++ {
dowork(n, i)
amt := time.Duration(rand.Intn(250))
time.Sleep(time.Millisecond * amt)
}
}
func dowork(goroutine, loopindex int) {
// simulate work
time.Sleep(time.Second * time.Duration(5))
fmt.Printf("gr[%d]: i=%d\n", goroutine, loopindex)
}
Can i assume that the 'dowork' function will be executed in parallel?
Is this a correct way of achieving parallelism or is it better to use channels and separate 'dowork' workers for each goroutine?

Answer 1

Regarding GOMAXPROCS, you can find this in Go 1.5's release docs:
By default, Go programs run with GOMAXPROCS set to the number of cores available; in prior releases it defaulted to 1.
Regarding preventing the main function from exiting immediately, you could leverage WaitGroup's Wait function.
I wrote this utility function to help parallelize a group of functions:
import "sync"
// Parallelize parallelizes the function calls
func Parallelize(functions ...func()) {
var waitGroup sync.WaitGroup
waitGroup.Add(len(functions))
defer waitGroup.Wait()
for _, function := range functions {
go func(copy func()) {
defer waitGroup.Done()
copy()
}(function)
}
}
So in your case, we could do this
func1 := func() {
f(0)
}
func2 = func() {
f(1)
}
func3 = func() {
f(2)
}
Parallelize(func1, func2, func3)
If you wanted to use the Parallelize function, you can find it here https://github.com/shomali11/util

Answer 2

This answer is outdated. Please see this answer instead.
Your code will run concurrently, but not in parallel. You can make it run in parallel by setting GOMAXPROCS.
It's not clear exactly what you're trying to accomplish here, but it looks like a perfectly valid way of achieving concurrency to me.

Answer 3

f() will be executed concurrently but many dowork() will be executed sequentially within each f(). Waiting on stdin is also not the right way to ensure that your routines finished execution. You must spin up a channel that each f() pushes a true on when the f() finishes.
At the end of the main() you must wait for n number of true's on the channel. n being the number of f() that you have spun up.

Answer 4

This helped me when I was starting out.
package main
import "fmt"
func put(number chan<- int, count int) {
i := 0
for ; i <= (5 * count); i++ {
number <- i
}
number <- -1
}
func subs(number chan<- int) {
i := 10
for ; i <= 19; i++ {
number <- i
}
}
func main() {
channel1 := make(chan int)
channel2 := make(chan int)
done := 0
sum := 0
go subs(channel2)
go put(channel1, 1)
go put(channel1, 2)
go put(channel1, 3)
go put(channel1, 4)
go put(channel1, 5)
for done != 5 {
select {
case elem := <-channel1:
if elem < 0 {
done++
} else {
sum += elem
fmt.Println(sum)
}
case sub := <-channel2:
sum -= sub
fmt.Printf("atimta : %d\n", sub)
fmt.Println(sum)
}
}
close(channel1)
close(channel2)
}
"Conventional cluster-based systems (such as supercomputers) employ parallel execution between processors using MPI. MPI is a communication interface between processes that execute in operating system instances on different processors; it doesn't support other process operations such as scheduling. (At the risk of complicating things further, because MPI processes are executed by operating systems, a single processor can run multiple MPI processes and/or a single MPI process can also execute multiple threads!)"

Answer 5

You can add a loop at the end, to block until the jobs are done:
package main
import "time"
func f(n int, b chan bool) {
println(n)
time.Sleep(time.Second)
b <- true
}
func main() {
b := make(chan bool, 9)
for n := cap(b); n > 0; n-- {
go f(n, b)
}
for <-b {
if len(b) == 0 { break }
}
}