golang functions: parallel execution with return

How to make two functions calls f1(2) and f1(1) execute in parallel so that all the program would execute for 2 seconds not for 3.

package main

import (
	"fmt"
	"time"
)

// sleeps for `secs` seconds
func f1(secs time.Duration) (result string) {
	fmt.Printf("waiting %V\n", secs)
	time.Sleep(secs * time.Second)
	result = fmt.Sprintf("waited for %d seconds", secs)
	return
}

// prints arg1, arg2
func f2(arg1, arg2 string) {
	fmt.Println(arg1)
	fmt.Println(arg2)
}

// this function executes for 3 seconds, because waits a lot
func runNotParallel() {

	out1 := f1(2)
	out2 := f1(1)
	f2(out1, out2)

}

// golang parallel return functions
// todo: make it run so all the function will executes for 2 seconds not for 3
func runParallel() {
	out1 := f1(2)
	out2 := f1(1)
	f2(out1, out2)
}

func main() {
	runNotParallel()
	runParallel()
}

playground

I guess I can do it only with channels. Should I redefine function f1 or I can leave it as is and change only way I call it?

Use chan/goroutine

package main

import (
	"fmt"
	"time"
)

// sleeps for `secs` seconds
func f1(secs time.Duration) (result string) {
	fmt.Printf("waiting %v\n", secs)
	time.Sleep(secs * time.Second)
	result = fmt.Sprintf("waited for %v seconds", secs)
	return
}

// prints arg1, arg2
func f2(arg1, arg2 string) {
	fmt.Println(arg1)
	fmt.Println(arg2)
}

// this function executes for 3 seconds, because waits a lot
func runNotParallel() {
	out1 := f1(2)
	out2 := f1(1)
	f2(out1, out2)

}

// golang parallel return functions
// todo: make it run so all the function will executes for 2 seconds not for 3
func runParallel() {
	out1 := make(chan string)
	out2 := make(chan string)
	go func() {
		out1 <- f1(2)
	}()
	go func() {
		out2 <- f1(1)
	}()
	f2(<-out1, <-out2)
}

func main() {
	runNotParallel()
	runParallel()
}

https://play.golang.org/p/G4RHiq9LJw

Another way you could do it is using WaitGroup

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

value1 := ""
value2 := ""

func1 := func() {
    value1 = f1(2)
}

func2 = func() {
    value2 = f1(1)
}

Parallelize(func1, func2)

f2(out1, out2)

If you wanted to use the Parallelize function, you can find it here https://github.com/shomali11/util

here is a solution without channels but with the missing f2 synchronization:

package main

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

// sleeps for `secs` seconds
func f1(secs time.Duration, result *string, sg *sync.WaitGroup) () {
	fmt.Printf("waiting %v\n", secs)
	time.Sleep(secs * time.Second)
	*result = fmt.Sprintf("waited for %d seconds", secs)
	if sg!= nil {
		sg.Done()
	}
	return
}

// prints arg1, arg2
func f2(arg1, arg2 string) {
	fmt.Println(arg1)
	fmt.Println(arg2)
}

// this function executes for 3 seconds, because waits a lot
func runNotParallel() {

	var out1, out2 string
	f1(2, &out1, nil)
	f1(1, &out2,nil)
	f2(out1, out2)

}

// golang parallel return functions
// todo: make it run so all the function will executes for 2 seconds not for 3
func runParallel() {
	var sg sync.WaitGroup
	sg.Add(2)
	var out1, out2 string
	go f1(2, &out1, &sg)
	go f1(1, &out2, &sg)
	sg.Wait()
	f2(out1, out2)
}

func main() {
	runNotParallel()
	runParallel()
}

basically, go operator blocks from using/accessing a return values but it could be done using a pointers for the return place holders in the signature

With go 1.18 supporting generics, the channel solution can be made even more readable.

func async[T any](f func() T) chan T {
	ch := make(chan T)
	go func() {
		ch <- f()
	}()
	return ch
}

func main() {
	startTime := time.Now().Local()

	out1 := async(func() string {
		time.Sleep(1 * time.Second)
		return "thing 1"
	})
	out2 := async(func() string {
		time.Sleep(2 * time.Second)
		return "thing 2"
	})

	results := []string{<-out1, <-out2}

	fmt.Printf("results: %v\n", results)
	fmt.Printf("took %v", time.Since(startTime))
}

playground

The lo package provides this along with lots of other generic helper functions.