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/* Copyright (C) 1992, 1997 Free Software Foundation, Inc.
This file is part of the GNU C Library.
* SPDXLicenseIdentifier: LGPL2.0+
*/
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;
}
