| /* 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_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; |
| } |