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+/* 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 <stdlib.h>
+
+
+/* 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;
+}