import newlib-2000-02-17 snapshot

1 files changed, 132 insertions, 0 deletions

diff --git a/newlib/libc/stdlib/div.c b/newlib/libc/stdlib/div.c
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+++ b/newlib/libc/stdlib/div.c
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+/*
+FUNCTION
+<<div>>---divide two integers
+
+INDEX
+	div
+
+ANSI_SYNOPSIS
+	#include <stdlib.h>
+	div_t div(int <[n]>, int <[d]>);
+
+TRAD_SYNOPSIS
+	#include <stdlib.h>
+	div_t div(<[n]>, <[d]>)
+	int <[n]>, <[d]>;
+
+DESCRIPTION
+Divide
+@tex
+$n/d$,
+@end tex
+@ifinfo
+<[n]>/<[d]>,
+@end ifinfo
+returning quotient and remainder as two integers in a structure <<div_t>>.
+
+RETURNS
+The result is represented with the structure
+
+. typedef struct
+. {
+.  int quot;
+.  int rem;
+. } div_t;
+
+where the <<quot>> field represents the quotient, and <<rem>> the
+remainder.  For nonzero <[d]>, if `<<<[r]> = div(<[n]>,<[d]>);>>' then
+<[n]> equals `<<<[r]>.rem + <[d]>*<[r]>.quot>>'.
+
+To divide <<long>> rather than <<int>> values, use the similar
+function <<ldiv>>.
+
+PORTABILITY
+<<div>> is ANSI.
+
+No supporting OS subroutines are required.
+*/
+
+/*
+ * Copyright (c) 1990 Regents of the University of California.
+ * All rights reserved.
+ *
+ * This code is derived from software contributed to Berkeley by
+ * Chris Torek.
+ *
+ * Redistribution and use in source and binary forms, with or without
+ * modification, are permitted provided that the following conditions
+ * are met:
+ * 1. Redistributions of source code must retain the above copyright
+ *    notice, this list of conditions and the following disclaimer.
+ * 2. Redistributions in binary form must reproduce the above copyright
+ *    notice, this list of conditions and the following disclaimer in the
+ *    documentation and/or other materials provided with the distribution.
+ * 3. All advertising materials mentioning features or use of this software
+ *    must display the following acknowledgement:
+ *	This product includes software developed by the University of
+ *	California, Berkeley and its contributors.
+ * 4. Neither the name of the University nor the names of its contributors
+ *    may be used to endorse or promote products derived from this software
+ *    without specific prior written permission.
+ *
+ * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
+ * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
+ * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
+ * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
+ * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
+ * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
+ * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
+ * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
+ * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
+ * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
+ * SUCH DAMAGE.
+ */
+
+#include <_ansi.h>
+#include <stdlib.h>		/* div_t */
+
+div_t
+_DEFUN (div, (num, denom),
+	int num _AND
+	int denom)
+{
+	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 or +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, but r.rem > 0, we also have the wrong answer.
+	 * In this case, to get the right answer, subtract 1 from r.quot and
+	 * add denom to r.rem.
+	 */
+	if (num >= 0 && r.rem < 0) {
+		++r.quot;
+		r.rem -= denom;
+	}
+	else if (num < 0 && r.rem > 0) {
+		--r.quot;
+		r.rem += denom;
+	}
+	return (r);
+}