| /* 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; | |
| } |