From 0ecb606cb6cf65de1d9fc8a919bceb4be476c602 Mon Sep 17 00:00:00 2001 From: Jakub Jelinek Date: Thu, 12 Jul 2007 18:26:36 +0000 Subject: 2.5-18.1 --- stdlib/ldiv.c | 54 ++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 54 insertions(+) create mode 100644 stdlib/ldiv.c (limited to 'stdlib/ldiv.c') diff --git a/stdlib/ldiv.c b/stdlib/ldiv.c new file mode 100644 index 0000000000..a7796d8e95 --- /dev/null +++ b/stdlib/ldiv.c @@ -0,0 +1,54 @@ +/* Copyright (C) 1992, 1997 Free Software Foundation, Inc. + This file is part of the GNU C Library. + + The GNU C Library is free software; you can redistribute it and/or + modify it under the terms of the GNU Lesser General Public + License as published by the Free Software Foundation; either + version 2.1 of the License, or (at your option) any later version. + + The GNU C Library is distributed in the hope that it will be useful, + but WITHOUT ANY WARRANTY; without even the implied warranty of + MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU + Lesser General Public License for more details. + + You should have received a copy of the GNU Lesser General Public + License along with the GNU C Library; if not, write to the Free + Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA + 02111-1307 USA. */ + +#include + + +/* Return the `ldiv_t' representation of NUMER over DENOM. */ +ldiv_t +ldiv (long int numer, long int denom) +{ + ldiv_t result; + + result.quot = numer / denom; + result.rem = numer % denom; + + /* The ANSI standard says that |QUOT| <= |NUMER / DENOM|, where + NUMER / DENOM is to be computed in infinite precision. In + other words, we should always truncate the quotient towards + zero, never -infinity. Machine division and remainer may + work either way when one or both of NUMER or DENOM is + negative. If only one is negative and QUOT has been + truncated towards -infinity, REM will have the same sign as + DENOM and the opposite sign of NUMER; if both are negative + and QUOT has been truncated towards -infinity, REM will be + positive (will have the opposite sign of NUMER). These are + considered `wrong'. If both are NUM and DENOM are positive, + RESULT will always be positive. This all boils down to: if + NUMER >= 0, but REM < 0, we got the wrong answer. In that + case, to get the right answer, add 1 to QUOT and subtract + DENOM from REM. */ + + if (numer >= 0 && result.rem < 0) + { + ++result.quot; + result.rem -= denom; + } + + return result; +} -- cgit v1.2.3