How to return values from goroutines?

Question

The below code gives compilation error saying 'unexpected go':
x := go doSomething(arg)

func doSomething(arg int) int{
    ...
    return my_int_value
}

I know, I can fetch the return value if call the function normally, without using goroutine. Or I can use channels etc.

My question is why is it not possible to fetch a return value like this from a goroutine.

Chris Gregori · Accepted Answer

The strict answer is that you can do that. It's just probably not a good idea. Here's code that would do that: var x int go func() { x = doSomething() }() This will spawn off a new goroutine which will calculate doSomething() and then assign the result to x. The problem is: how are you going to use x from the original goroutine? You probably want to make sure the spawned goroutine is done with it so that you don't have a race condition. But if you want to do that, you'll need a way to communicate with the goroutine, and if you've got a way to do that, why not just use it to send the value back? OR The most natural way to fetch a value from a goroutine is channels. Channels are the pipes that connect concurrent goroutines. You can send values into channels from one goroutine and receive those values into another goroutine or in a synchronous function. You could easily obtain a value from a goroutine not breaking concurrency using select: package main import ( "fmt" "time" ) func main() { c1 := make(chan string) c2 := make(chan string) go func() { time.Sleep(time.Second * 1) c1 <- "one" }() go func() { time.Sleep(time.Second * 2) c2 <- "two" }() for i := 0; i < 2; i++ { // Await both of these values // simultaneously, printing each one as it arrives. select { case msg1 := <-c1: fmt.Println("received", msg1) case msg2 := <-c2: fmt.Println("received", msg2) } } } https://play.golang.org/p/_MS2nU2m5Nx

Go Language

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