Golang Timers with 0 length - go

Question

I've written a code snipped that creates a timer with a 0 length time, and it does not immediately expire (which is what I expected). A very short sleep call does make it expire, but I'm confused as to why.
The reason I care is that the code using this idea has a snippet that returns 0 on a low probability error, with the idea that the timer should be set to immediately expire, and retry a function. I do not believe that the nanosecond sleep needed here will affect my implementation, but it bothers me.
Did I make a mistake, is this expected behaviour?
Thanks!
package main
import (
"fmt"
"time"
)
func main() {
testTimer := time.NewTimer(time.Duration(0) * time.Millisecond)
fmt.Println(Expired(testTimer))
time.Sleep(time.Nanosecond)
fmt.Println(Expired(testTimer))
}
func Expired(T *time.Timer) bool {
select {
case <-T.C:
return true
default:
return false
}
}
Playground link: https://play.golang.org/p/xLLHoR8aKq
Prints
false
true

Answer 1

time.NewTimer() does not guarantee maximum wait time. It only guarantees a minimum wait time. Quoting from its doc:
NewTimer creates a new Timer that will send the current time on its channel after at least duration d.
So passing a zero duration to time.NewTimer(), it's not a surprise the returned time.Timer is not "expired" immediately.
The returned timer could be "expired" immediately if the implementation would check if the passed duration is zero, and would send a value on the timer's channel before returning it, but it does not. Instead it starts an internal timer normally as it does for any given duration, which will take care of sending a value on its channel, but only some time in the future.
Note that with multiple CPU cores and with runtime.GOMAXPROCS() being greater than 1 there is a slight chance that another goroutine (internal to the time package) sends a value on the timer's channel before NewTimer() returns, but this is a very small chance... Also since this is implementation detail, a future version might add this "optimization" to check for 0 passed duration, and act as you expected it, but as with all implementation details, don't count on it. Count on what's documented, and expect no more.

Answer 2

Go's timer functions guarantee to sleep at least the specified time. See the docs for Sleep and NewTimer respectively:
Sleep pauses the current goroutine for at least the duration d. A negative or zero duration causes Sleep to return immediately.
NewTimer creates a new Timer that will send the current time on its channel after at least duration d.
(emphasis added)
In your situation, you should probably just not use a timer in the situation that you don't want to sleep at all.

Answer 3

This is due to the internal time it takes to set up the timer object. If you'll note in the playground link below the timer does expire at the proper time, but the internal go routine that sets it up and starts it takes longer than your Expire function does to check it.
When the Timer expires, the current time will be sent on C (the channel)
So you'll notice that after it expires, it still sends the original time, because it has expired even before the nanosecond Sleep finished.
https://play.golang.org/p/Ghwq9kJq3J
package main
import (
"fmt"
"time"
)
func main() {
testTimer := time.NewTimer(0 * time.Millisecond)
Expired(testTimer)
time.Sleep(time.Nanosecond)
Expired(testTimer)
n := time.Now()
fmt.Printf("after waiting: %d\n", n.UnixNano())
}
func Expired(T *time.Timer) bool {
select {
case t:= <-T.C:
fmt.Printf("expired %d\n", t.UnixNano())
return true
default:
n := time.Now()
fmt.Printf("not expired: %d\n", n.UnixNano())
return false
}
}