runtime: avoid runtimeNano call on a common netpoll path

runtimeNano is slower than nanotime, so pass the duration
to runtime_pollSetDeadline as is. netpoll can add nanotime itself.
Arguably a bit simpler because, say, a negative duration
clearly represents already expired timer, no need to compare to
nanotime again.
This may also fix an obscure corner case when a deadline in past
which happens to be nanotime 0 is confused with no deadline at all,
which are radically different things.
Also don't compute any durations and times if Time is zero
(currently we first compute everything and then reset d back to 0,
which is wasteful).

name                  old time/op  new time/op  delta
TCP4OneShotTimeout-6  17.1µs ± 0%  17.0µs ± 0%     ~     (p=0.421 n=5+5)
SetReadDeadline-6      230ns ± 0%   205ns ± 1%  -10.63%  (p=0.008 n=5+5)

Change-Id: I2aad699270289a5b9ead68f5e44ec4ec6d96baa0
Reviewed-on: https://go-review.googlesource.com/c/146344
Reviewed-by: Ian Lance Taylor <iant@golang.org>
Run-TryBot: Dmitry Vyukov <dvyukov@google.com>

src/internal/poll/fd_poll_runtime.go

@@ -136,15 +136,12 @@ func (fd *FD) SetWriteDeadline(t time.Time) error {
 }
 
 func setDeadlineImpl(fd *FD, t time.Time, mode int) error {
-	diff := int64(time.Until(t))
-	d := runtimeNano() + diff
-	if d <= 0 && diff > 0 {
-		// If the user has a deadline in the future, but the delay calculation
-		// overflows, then set the deadline to the maximum possible value.
-		d = 1<<63 - 1
+	var d int64
+	if !t.IsZero() {
+		d = int64(time.Until(t))
+		if d == 0 {
+			d = -1 // don't confuse deadline right now with no deadline
 		}
-	if t.IsZero() {
-		d = 0
 	}
 	if err := fd.incref(); err != nil {
 		return err

src/runtime/netpoll.go

@@ -201,8 +201,13 @@ func poll_runtime_pollSetDeadline(pd *pollDesc, d int64, mode int) {
 	}
 	rd0, wd0 := pd.rd, pd.wd
 	combo0 := rd0 > 0 && rd0 == wd0
-	if d != 0 && d <= nanotime() {
-		d = -1
+	if d > 0 {
+		d += nanotime()
+		if d <= 0 {
+			// If the user has a deadline in the future, but the delay calculation
+			// overflows, then set the deadline to the maximum possible value.
+			d = 1<<63 - 1
+		}
 	}
 	if mode == 'r' || mode == 'r'+'w' {
 		pd.rd = d