How to use a CHANNEL? Golang always tell me throw: all goroutines are asleep - deadlock!

[feeling4t]

Hi guys, I'm really new in golang

when I try this code:

func main(){

ch := make(chan int)

ch <- 2

i :=<- ch

fmt.Printf("%d",i)

}

the golang always tell me

throw: all goroutines are asleep - deadlock!

goroutine 1 [chan send]:

main.main()

/home/feeling4t/workspace/routetest/src/main.go:16 +0x51

goroutine 2 [syscall]:

created by runtime.main

/tmp/bindist046461602/go/src/pkg/runtime/proc.c:221

I think the way I use channel must be wrong, could any one tell me why?

Thanks

[Daniël Bos (远洋)]

As Peter Russell already mentioned, the channel you are using is
unbuffered, so since no-one is listening at the moment when you write
to the channel, the call will block.

Usually channels are used to communicate between different threads of
operation (go-routines), you could use it like this:

func main(){
ch := make(chan int)

go func() { ch <- 2 }()

i :=<- ch
fmt.Printf("%d",i)
}

This will create an anonymous function that will write to the channel,
and execute it in a separate thread. Because this runs in parallel,
the following line, where you read from the channel, is reached, which
will un-block the writer, and all is well.

You can play with it here: http://play.golang.org/p/macbX7u2up

--
远洋 / Daniël Bos

[feeling4t]

It did works

Thanks

在 2012年7月24日星期二UTC+8上午11时24分07秒，Peter Russell写道：

> I think the way I use channel must be wrong, could any one tell me why?

Because the channel is unbuffered, the sending goroutine blocks until a goroutine receives a value from the channel. There is no other goroutine, so it's a deadlock.

Try this: http://play.golang.org/p/1agVxICOZ3

I converted the unbuffered channel to a buffered channel and I think it works the way you expect.

[feeling4t]

This makes me clear, Thank U : )

在 2012年7月24日星期二UTC+8上午11时50分54秒，Daniel Bos写道：