error for using len() function to control a channel

[Yang-YiFan]

Please answer these questions before submitting your issue. Thanks!

What version of Go are you using (go version)?

go1.9.1 linux/amd64

Does this issue reproduce with the latest release?

yes

What operating system and processor architecture are you using (go env)?

GOARCH="amd64"
GOBIN=""
GOEXE=""
GOHOSTARCH="amd64"
GOHOSTOS="linux"
GOOS="linux"
GOPATH="/home/yangyf/go"
GORACE=""
GOROOT="/usr/local/go"
GOTOOLDIR="/usr/local/go/pkg/tool/linux_amd64"
GCCGO="gccgo"
CC="gcc"
GOGCCFLAGS="-fPIC -m64 -pthread -fmessage-length=0 -fdebug-prefix-map=/tmp/go-build419888304=/tmp/go-build -gno-record-gcc-switches"
CXX="g++"
CGO_ENABLED="1"
CGO_CFLAGS="-g -O2"
CGO_CPPFLAGS=""
CGO_CXXFLAGS="-g -O2"
CGO_FFLAGS="-g -O2"
CGO_LDFLAGS="-g -O2"
PKG_CONFIG="pkg-config"

What did you do?

If possible, provide a recipe for reproducing the error.
A complete runnable program is good.
A link on play.golang.org is best.

just a very simple Go program

package main
import "fmt"
import "time"

func main() {
	a:=make(chan int,10)
	go func() {
		for {
			if(len(a)!=0) {
				x:=<-a
				fmt.Println(x)
				if x==-1 {
					break
				}
			}
		}
	}()

	time.Sleep(2*time.Second)
	a<-1
	a<-2
	a<-3
	time.Sleep(2*time.Second)
	a<-6
	a<-7
	a<- -1
        time.Sleep(5*time.Second)
}

What did you expect to see?

the program should behave as if the line if (len(a))!=0 doesn't exist
that is channel a receive the data sent from main function and print it
it will print 1,2,3,6,7,-1 respectively

What did you see instead?

instead there is no print at all
if I print len(a), it will constantly show len(a)=0 rather than some positive value
in some sense channel a is not working? I don't know why
so the program can't stop

[ianlancetaylor]

Closing as dup of #10958.

It's unfortunate, and we're working on it. But also, don't write programs this way. This program will burn an entire CPU core doing nothing.

[heschi]

(If you got a notification for a previous comment from me, please ignore, I meant to put that on #22434.)