According to https://golang.org/pkg/time/#Duration.Round "If m <= 0, Round returns d unchanged". So now I'm really curious, why do the Round() at all?
s On Sunday, April 19, 2020 at 6:45:45 PM UTC-7, ise...@gmail.com wrote: > > Hi, > > I have been looking at various packages to understand how one goes about > implementing things in Go. I have been learning a great deal just by > getting my head around code written by smarter devs than I'm ever likely to > be ;) > > Whilst looking at golang/oauth2, two lines of code handling token expiry > led me to discover about time.Time's use of both wall and monotonic clocks. > I realised that either I had failed to understand something or the code was > wrong. I'm assuming it's the former, so hopefully somebody can explain it > to me... > > In oauth2/internal/token.go, doTokenRoudnTrip sets the Token.Expiry as > follows: > > token.Expiry = time.Now().Add(time.Duration(expires) * time.Second) > > In oauth2/token.go, the Token.expired function returns this: > > return t.Expiry.Round(0).Add(-expiryDelta).Before(timeNow()) > > If I have read the time package documentation correctly, the monotonic > clock reading is intended for time measurements. In particular, it provides > robust behaviour for functions such as Before(). > > So I do not understand why the monotonic reading is stripped from the time > value with Round(0) before invoking Before() in the code above. As I read > it, the token.Expiry has a monotonic clock reading (now + duration) which > is then stripped when evaluated against Before(now) later on... does this > not make the code susceptible to issues when system time is being adjusted? > What is the rationale for Round(0) in this case? > > -- ise > > > > > -- You received this message because you are subscribed to the Google Groups "golang-nuts" group. To unsubscribe from this group and stop receiving emails from it, send an email to golang-nuts+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/golang-nuts/a4167638-c04a-4e64-9374-67fe240135c4%40googlegroups.com.
