| /* 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; |

| } |