CODEWITHSUNDEEP
go
Arpssss
Arpssss

Reputation: 3858

Go Programming Language Code Error

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
        }

}

Upvotes: 0

Views: 222

Answers (2)

David Grayson
David Grayson

Reputation: 87376

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

Upvotes: 2

Matt K
Matt K

Reputation: 13832

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.

Upvotes: 2

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