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• Go channels and deadlock 2 answers

I am trying to have two separate consumer go routines, that would filter out even and odd numbers from the input channel. This is just a toy example in order to see if it is possible to have consumers do something with the message read from the input channel if it matches certain condition, otherwise put back onto the input channel.

My current code is as follows:

package main

func filterOdd(ch chan int, out chan int) {
    val := <- ch
    if val % 2 == 0 {
        ch <- val
    } else {
        out <- val
    }
}
func filterEven(ch chan int, out chan int) {
    val := <- ch
    if val % 2 != 0 {
        ch <- val
    } else {
        out <- val
    }
}

func main() {
    even := make(chan int)
    odd := make(chan int)
    input := make(chan int)
    go filterOdd(input, odd)
    go filterEven(input, even)
    for i:=1; i <= 10; i++ {
        input <- i
    }

    println("Even...")
    for i := range even {
        println(i)
    }

    println("Odd...")
    for i := range odd {
        println(i)
    }
}

However, this produces the following output:

fatal error: all goroutines are asleep - deadlock!

goroutine 1 [chan send]:
main.main()
    /tmp/sandbox594577124/main.go:27 +0x140

goroutine 4 [chan send]:
main.filterOdd(0x10336100, 0x103360c0)
    /tmp/sandbox594577124/main.go:8 +0xc0
created by main.main
    /tmp/sandbox594577124/main.go:24 +0xc0

Link to the Go Playground: https://play.golang.org/p/9RIvFsGKI-

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Answer 1

You have a deadlock because your even and odd goroutines are blocked on sending to out because nothing is reading from it. Why is nothing reading out? Because the main goroutine is blocked on sending to input because nothing is reading from it. Why is nothing reading from input? Because the two goroutines that would read from it are blocked.

Also, both filterEven and filterOdd will only run once unless you wrap their contents in for { } (but then they'll never stop until you break). On the other hand, range even will block (and range odd never happens) when nothing is left to write to even , because range over a channel only stops when the channel is closed or break is called.

In general, these aren't hard problems to solve as long as you know when you can close a channel. With what you're describing, that gets more difficult. None of the goroutines know when it's OK to close input, because all three write to it and two also read from it. You can use a sync.WaitGroup to make sure that everything you've put into the input channel has been processed before closing it. Once it's closed, the two other goroutines can use that as a signal to close their own channels and break or return to finish running.

However, writes to the in and out channels will still block until there is a corresponding read, because they are unbuffered. However, if you buffer them by specifying a size as the second argument to make, writes won't block until the channel is full. Since you know neither even or odd will have more written to them than what main send to input, you can use that as a safe buffer capacity.

Here's an example of using a WaitGroup with buffered channels for your code: https://play.golang.org/p/VXqfwUwRcx

If you don't want buffered channels, you can also use another pair of goroutines to capture the values and send them back to main as slices once finished. This way writes on the even and odd channels don't block: https://play.golang.org/p/i5vLDcsK1v

Otherwise, if don't need to print the contents of each channel all at once, you can use those two extra goroutines to read from the channels and print right away: https://play.golang.org/p/OCaUTcJkKB