blob: 8e11333acdf391cd9113afa62e2ba6cce8d9ccfb [file] [log] [blame]
/* Copyright (C) 1992, 1997 Free Software Foundation, Inc.
This file is part of the GNU C Library.
* SPDX-License-Identifier: LGPL-2.0+
typedef struct {
long quot;
long rem;
} ldiv_t;
/* Return the `ldiv_t' representation of NUMER over DENOM. */
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.rem -= denom;
return result;