Go Programming Language Code Error

Question

I try to write a simple code in GO, in which two go routine (Send and Receive) send each other integers. I give the code below. Can anybody help me why the output of this program is [no output]. Is there any silly mistake (sorry, I am new in GO) ?

package main

func Send (in1 <-chan int, out1 chan<- int) {

                i := 2 
        out1 <- i 
    print(i, "\n")
}

func Receive (in <-chan int, out chan<- int) {

        i := <-in 
        print(i, "\n")
        out <- i 
}


func main() {

       for i := 0; i < 10; i++ {
        ch1 := make(chan int)
        ch := make(chan int) 
            go Send (ch1 , ch)
        go Receive (ch , ch1)
        ch = ch1
        ch1 = ch
        }

}

Fix your indentation before posting.

David Grayson
– David Grayson

2011-11-22 16:13:51 +00:00
Commented Nov 22, 2011 at 16:13 — David Grayson
– David Grayson, Commented Nov 22, 2011 at 16:13

David Grayson · Accepted Answer · 2011-11-22 17:09:34Z

2

How about this:

package main

func Send (ch chan<- int) {
    for i := 0; i < 10; i++ {
      print(i, " sending\n")
      ch <- i
    }
}

func Receive (ch <-chan int) {
    for i := 0; i < 10; i++ {
        print(<-ch, " received\n")
    }
}

func main() {
   ch := make(chan int)
   go Receive(ch)
   Send(ch)
}

The output of this when I run it on golang.org is:

0 sending
0 received
1 sending
2 sending
1 received
2 received
3 sending
4 sending
3 received
4 received
5 sending
6 sending
5 received
6 received
7 sending
8 sending
7 received
8 received
9 sending

I'm not sure why the 9 was never received. There should be some way to sleep the main thread until the Receive goroutine has completed. Also, it's inelegant that the receiver and the sender both know that they are going to send 10 numbers. One of the goroutines should just shut off when the other has finished its work. I'm not sure how to do that.

EDIT1:

Here is an implementation with two channels, and the go routines send ints back and forth between eachother. One is designated as the responder, who only sends and int after it receives an int. The responder simply adds two to the int he receives, and then sends it back.

package main

func Commander(commands chan int, responses chan int) {
    for i := 0; i < 10; i++ {
      print(i, " command\n")
      commands <- i
      print(<-responses, " response\n");
    }
    close(commands)
}

func Responder(commands chan int, responses chan int) {
    for {
        x, open := <-commands
        if !open {
            return;
        }
        responses <- x + 2
    }
}

func main() {
   commands := make(chan int)
   responses := make(chan int)
   go Commander(commands, responses)
   Responder(commands, responses)
}

The output when I run it on golang.org is:

0 command
2 response
1 command
3 response
2 command
4 response
3 command
5 response
4 command
6 response
5 command
7 response
6 command
8 response
7 command
9 response
8 command
10 response
9 command
11 response

edited Nov 22, 2011 at 17:09

answered Nov 22, 2011 at 16:13

David Grayson

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5 Comments

Arpssss Over a year ago

Thanks David Grayson. It's cool.. However, what I actually want to do Send routine sends int to receive routine and also receives int from it. Means two way communication channel. Is it possible ?

Matt K Over a year ago

The send routine should close the channel after sending the last integer, and the receiver should check for a closed channel when receiving new values from the channel: golang.org/doc/go_spec.html#Close

Matt K Over a year ago

@Arpssss: you need two channels for that.

Matt K Over a year ago

Feel free to incorporate my answer. All that stuff wouldn't fit into a comment here :)

David Grayson Over a year ago

Okay! Please see EDIT1 in my answer; I have incorporated your comments, Arpssss and @mkb.

Matt K · Accepted Answer · 2011-11-22 16:44:44Z

David Grayson's fix works, but the explanation of why 9 is not always received wouldn't fit into a comment!

When the sending goroutine sends the value on the channel, it blocks until the receiving goroutine receives it. At that point, it's unblocked. The go scheduler might return from main almost immediately before it gives the receiving goroutine a chance to print anything. If you set GOMAXPROCS to something greater than 1, this happens less frequently because the receiving goroutine won't block while the main goroutine is active.

If you want to stop main from returning until all goroutines are finished, you can return a channel from Receive(), like this:

package main

import "fmt"

func Send(ch chan<- int) {
    for i := 0; i < 10; i++ {
        fmt.Println(i, " sending")
        ch <- i
    }
    close(ch)
}

func Receive(ch <-chan int) (<-chan bool) {
    done := make(chan bool)
    go func() {
        for {
            i, ok := <-ch
            if !ok {
                break
            }
            fmt.Println(i, " received")
        }
        done <- true
    }()
    return done
}

func main() {
    ch := make(chan int)
    d := Receive(ch)
    Send(ch)
    _ = <-d
}

As for why your output was empty in your original example: you make two goroutines, and thus main is totally unblocked, so it just returns and the program exits! I've definitely made this mistake too, but the runtime only joins with the main goroutine, not with ALL goroutines, like you may expect.

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