golang函数：带返回的并行执行

Question

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
", 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?

Answer 1

Use chan/goroutine

package main

import (
    "fmt"
    "time"
)

// sleeps for `secs` seconds
func f1(secs time.Duration) (result string) {
    fmt.Printf("waiting %v
", 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