/* FUNCTION <
>---divide two integers INDEX div ANSI_SYNOPSIS #include div_t div(int <[n]>, int <[d]>); TRAD_SYNOPSIS #include div_t div(<[n]>, <[d]>) int <[n]>, <[d]>; DESCRIPTION Divide @tex $n/d$, @end tex @ifnottex <[n]>/<[d]>, @end ifnottex returning quotient and remainder as two integers in a structure <>. RETURNS The result is represented with the structure . typedef struct . { . int quot; . int rem; . } div_t; where the <> field represents the quotient, and <> the remainder. For nonzero <[d]>, if `<<<[r]> = div(<[n]>,<[d]>);>>' then <[n]> equals `<<<[r]>.rem + <[d]>*<[r]>.quot>>'. To divide <> rather than <> values, use the similar function <>. PORTABILITY <
> 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 /* 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); }