43 lines
1.4 KiB
C
43 lines
1.4 KiB
C
// SPDX-License-Identifier: GPL-2.0+
|
|
/* Copyright (C) 1992, 1997 Free Software Foundation, Inc.
|
|
This file is part of the GNU C Library.
|
|
*/
|
|
|
|
typedef struct {
|
|
long quot;
|
|
long rem;
|
|
} ldiv_t;
|
|
/* 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;
|
|
}
|