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Friday, September 30, 2022

[FIXED] Why do I have a deadlock even though it includes an endless loop?

 September 30, 2022     concurrency, deadlock, go, multiprocessing     No comments   

Issue

I've started messing around with Go, and I'm fascinated by go routines. I now wrote a simple test to see if I can change the value of a variable while I continuously print it out.

I now have the following code:

package main

import (
    "fmt"
    "time"
)

func change(c chan float64) float64 {
    time.Sleep(2 * time.Second)
    return 2.5
}

func main() {

    s := 1.1

    c := make(chan float64)
    go change(c)
    s = <-c

    for {
        fmt.Println(s)
        time.Sleep(100 * time.Millisecond)
    }    
}

Unfortunately it ends in an error:

fatal error: all goroutines are asleep - deadlock!

goroutine 1 [chan receive]:
main.main()
        /home/kramer65/repos/go/src/messing_around/main.go:19 +0x7d
exit status 2

I searched around and found that this deadlock happens when the main function ends while goroutines didn't end yet. But since I have an endless loop I wouldn't know what else is wrong with my code.

What is wrong with this code, and how I can change the value of a variable while I continuously print it out?


Solution

It looks like you might have a misunderstanding on channels and go routines.

The line:

go change(c)

seems to indicate that the function change is going to write to c. However it ends up just returning a value after a period of time.

This value (2.5) doesn't get received anywhere. Also, c doesn't get written to anywhere. I suspect you intended the 2.5 to be written to the channel c. The syntax for that is as follows:

c<-2.5

Therefore, if you change your change function to:

func change(c chan float64) {
    time.Sleep(2 * time.Second)
    c <- 2.5
}

You should not see the deadlock anymore. Notice I am not returning a float64 anymore.

I made a playground to ensure this: https://play.golang.org/p/SgLiUmPpcAZ

Update for Comment

The 1.1 will always be overwritten by the value of the channel. If however you want to print s's initial value (as stated in the comment), you will have to change the flow a bit and use a select statement:

package main

import (
    "fmt"
    "time"
)

func change(c chan float64) {
    time.Sleep(2 * time.Second)
    c <- 2.5
}

func main() {

    s := 1.1

    c := make(chan float64)
    go change(c)

    for {
        select {
        case s = <-c:
        default:
            // c isn't ready yet
        }
        fmt.Println(s)
        time.Sleep(100 * time.Millisecond)
    }
}

Now that you have a select statement, you can use it with a time.Ticker as well:

ticker := time.NewTicker(100 * time.Millisecond)
for {
    select {
    case s = <-c:
    case <-ticker.C:
        fmt.Println(s)
    default:
        // c isn't ready yet
    }
}


Answered By - poy
Answer Checked By - Clifford M. (PHPFixing Volunteer)
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