1 files changed, 34 insertions, 34 deletions

diff --git a/main/glib/div.c b/main/glib/div.c
index 7dfe553..4384cc0 100644
--- a/main/glib/div.c
+++ b/main/glib/div.c
@@ -1,6 +1,6 @@
 /*
  * Copyright (c) 1990, 1993
- *	The Regents of the University of California.  All rights reserved.
+ *  The Regents of the University of California.  All rights reserved.
  *
  * This code is derived from software contributed to Berkeley by
  * Chris Torek.
@@ -36,42 +36,42 @@ static char sccsid[] = "@(#)div.c	8.1 (Berkeley) 6/4/93";
 #include <sys/cdefs.h>
 __FBSDID("$FreeBSD$");
 
-#include <stdlib.h>		/* div_t */
+#include <stdlib.h>     /* div_t */
 
 div_t
 div(num, denom)
-	int num, denom;
+int num, denom;
 {
-	div_t r;
+    div_t r;
 
-	r.quot = num / denom;
-	r.rem = num % denom;
-	/*
-	 * The ANSI standard says that |r.quot| <= |n/d|, where
-	 * n/d is to be computed in infinite precision.  In other
-	 * words, we should always truncate the quotient towards
-	 * 0, never -infinity.
-	 *
-	 * Machine division and remainer may work either way when
-	 * one or both of n or d is negative.  If only one is
-	 * negative and r.quot has been truncated towards -inf,
-	 * r.rem will have the same sign as denom and the opposite
-	 * sign of num; if both are negative and r.quot has been
-	 * truncated towards -inf, r.rem will be positive (will
-	 * have the opposite sign of num).  These are considered
-	 * `wrong'.
-	 *
-	 * If both are num and denom are positive, r will always
-	 * be positive.
-	 *
-	 * This all boils down to:
-	 *	if num >= 0, but r.rem < 0, we got the wrong answer.
-	 * In that case, to get the right answer, add 1 to r.quot and
-	 * subtract denom from r.rem.
-	 */
-	if (num >= 0 && r.rem < 0) {
-		r.quot++;
-		r.rem -= denom;
-	}
-	return (r);
+    r.quot = num / denom;
+    r.rem = num % denom;
+    /*
+     * The ANSI standard says that |r.quot| <= |n/d|, where
+     * n/d is to be computed in infinite precision.  In other
+     * words, we should always truncate the quotient towards
+     * 0, never -infinity.
+     *
+     * Machine division and remainer may work either way when
+     * one or both of n or d is negative.  If only one is
+     * negative and r.quot has been truncated towards -inf,
+     * r.rem will have the same sign as denom and the opposite
+     * sign of num; if both are negative and r.quot has been
+     * truncated towards -inf, r.rem will be positive (will
+     * have the opposite sign of num).  These are considered
+     * `wrong'.
+     *
+     * If both are num and denom are positive, r will always
+     * be positive.
+     *
+     * This all boils down to:
+     *  if num >= 0, but r.rem < 0, we got the wrong answer.
+     * In that case, to get the right answer, add 1 to r.quot and
+     * subtract denom from r.rem.
+     */
+    if(num >= 0 && r.rem < 0) {
+        r.quot++;
+        r.rem -= denom;
+    }
+    return (r);
 }