/* `long long int' divison with remainder.
- Copyright (C) 1992, 1996, 1997 Free Software Foundation, Inc.
+ Copyright (C) 1992-2018 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
/* Return the `lldiv_t' representation of NUMER over DENOM. */
lldiv_t
-lldiv (numer, denom)
- long long int numer;
- long long int denom;
+lldiv (long long int numer, long long int denom)
{
lldiv_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;
}