1 files changed, 51 insertions, 1 deletions
diff --git a/sys/kern/kern_clock.c b/sys/kern/kern_clock.c
index 0b7dfd3fe9e..6e6f0e3cca5 100644
--- a/sys/kern/kern_clock.c
+++ b/sys/kern/kern_clock.c
@@ -1,4 +1,4 @@
-/*	$OpenBSD: kern_clock.c,v 1.31 2002/01/02 06:07:41 nordin Exp $	*/
+/*	$OpenBSD: kern_clock.c,v 1.32 2002/02/15 01:59:26 art Exp $	*/
 /*	$NetBSD: kern_clock.c,v 1.34 1996/06/09 04:51:03 briggs Exp $	*/
 
 /*-
@@ -764,6 +764,56 @@ hzto(tv)
 }
 
 /*
+ * Compute number of hz in the specified amount of time.
+ */
+int
+tvtohz(struct timeval *tv)
+{
+	unsigned long ticks;
+	long sec, usec;
+
+	/*
+	 * If the number of usecs in the whole seconds part of the time
+	 * difference fits in a long, then the total number of usecs will
+	 * fit in an unsigned long.  Compute the total and convert it to
+	 * ticks, rounding up and adding 1 to allow for the current tick
+	 * to expire.  Rounding also depends on unsigned long arithmetic
+	 * to avoid overflow.
+	 *
+	 * Otherwise, if the number of ticks in the whole seconds part of
+	 * the time difference fits in a long, then convert the parts to
+	 * ticks separately and add, using similar rounding methods and
+	 * overflow avoidance.  This method would work in the previous
+	 * case but it is slightly slower and assumes that hz is integral.
+	 *
+	 * Otherwise, round the time difference down to the maximum
+	 * representable value.
+	 *
+	 * If ints have 32 bits, then the maximum value for any timeout in
+	 * 10ms ticks is 248 days.
+	 */
+	sec = tv->tv_sec;
+	usec = tv->tv_usec;
+	if (usec < 0) {
+		sec--;
+		usec += 1000000;
+	}
+	if (sec < 0 || (sec == 0 && usec <= 0)) {
+		ticks = 0;
+	} else if (sec <= LONG_MAX / 1000000)
+		ticks = (sec * 1000000 + (unsigned long)usec + (tick - 1))
+		    / tick + 1;
+	else if (sec <= LONG_MAX / hz)
+		ticks = sec * hz
+		    + ((unsigned long)usec + (tick - 1)) / tick + 1;
+	else
+		ticks = LONG_MAX;
+	if (ticks > INT_MAX)
+		ticks = INT_MAX;
+	return ((int)ticks);
+}
+
+/*
  * Start profiling on a process.
  *
  * Kernel profiling passes proc0 which never exits and hence