Page 253 of The Go Programming Language states:
... if instead of returning from
main
in the event of cancellation, we execute a call topanic
, then the runtime will dump the stack of every goroutine in the program.
This code deliberately leaks a goroutine by waiting on a channel that never has anything to receive:
package main
import (
"fmt"
"time"
)
func main() {
never := make(chan struct{})
go func() {
defer fmt.Println("End of child")
<-never
}()
time.Sleep(10 * time.Second)
panic("End of main")
}
However, the runtime only lists the main goroutine when panic
is called:
panic: End of main
goroutine 1 [running]:
main.main()
/home/simon/panic/main.go:15 +0x7f
exit status 2
If I press Ctrl-\ to send SIGQUIT during the ten seconds before main
panics, I do see the child goroutine listed in the output:
goroutine 1 [sleep]:
time.Sleep(0x2540be400)
/usr/lib/go-1.17/src/runtime/time.go:193 +0x12e
main.main()
/home/simon/panic/main.go:14 +0x6c
goroutine 18 [chan receive]:
main.main.func1()
/home/simon/panic/main.go:12 +0x76
created by main.main
/home/simon/panic/main.go:10 +0x5d
I thought maybe the channel was getting closed as panic
runs (which still wouldn't guarantee the deferred fmt.Println
had time to execute), but I get the same behaviour if the child goroutine does a time.Sleep
instead of waiting on a channel.
I know there are ways to dump goroutine stacktraces myself, but my question is why doesn't panic
behave as described in the book? The language spec only says that a panic will terminate the program, so is the book simply describing implementation-dependent behaviour?