--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/genericopenlibs/cstdlib/LSTDLIB/DIV.C	Fri Jun 04 16:20:51 2010 +0100
@@ -0,0 +1,122 @@
+/*
+* 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]>;
+* 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 */
+
+/**
+Divide two integer values.
+numer is divided by denom. 
+Quotient and remainder are returned in a div_t structure.
+@return   A div_t structure is returned
+@param num Numerator. 
+@param denom Denominator.
+*/
+EXPORT_C div_t div (int num, 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.
+	 *
+	 * 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);
+}