whitespace

diff --git a/usr.bin/cal/README b/usr.bin/cal/README
index 638ac9d..8305165 100644 (file)
--- a/usr.bin/cal/README
+++ b/usr.bin/cal/README
@@ -10,13 +10,13 @@ that even if I was dumb enough to try.  From this we can easily calculate
 the day of week for any date.  The algorithm for a zero based day of week:
 
        calculate the number of days in all prior years (year-1)*365
-       add the number of leap years (days?) since year 1 
+       add the number of leap years (days?) since year 1
                (not including this year as that is covered later)
        add the day number within the year
                this compensates for the non-inclusive leap year
                calculation
        if the day in question occurs before the gregorian reformation
-               (3 sep 1752 for our purposes), then simply return 
+               (3 sep 1752 for our purposes), then simply return
                (value so far - 1 + SATURDAY's value of 6) modulo 7.
        if the day in question occurs during the reformation (3 sep 1752
                to 13 sep 1752 inclusive) return THURSDAY. This is my

diff --git a/usr.bin/cal/cal.c b/usr.bin/cal/cal.c
index c37402d..ea6c812 100644 (file)
--- a/usr.bin/cal/cal.c
+++ b/usr.bin/cal/cal.c
@@ -1,4 +1,4 @@
-/*     $OpenBSD: cal.c,v 1.31 2022/12/04 23:50:47 cheloha Exp $        */
+/*     $OpenBSD: cal.c,v 1.32 2024/08/18 19:58:35 deraadt Exp $        */
 /*     $NetBSD: cal.c,v 1.6 1995/03/26 03:10:24 glass Exp $    */
 
 /*
@@ -261,7 +261,7 @@ week(int day, int month, int year)
        shift = 1;
        if (yearday < firstsunday)
                return (1);
-       if (firstweekday > THURSDAY - 1) 
+       if (firstweekday > THURSDAY - 1)
                shift = 2;
        return ((((yearday + 1) - (weekday - 1)) / 7) + shift);
 }
@@ -291,7 +291,7 @@ isoweek(int day, int month, int year)
                return 53 - (g - s) / 5;
        else if (n > 364 + s)
                return 1;
-       else 
+       else
                return n/7 + 1;
 }