/* Copyright (C) 1992, 1997 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 modify it under the terms of the GNU Library General Public License as published by the Free Software Foundation; either version 2 of the License, or (at your option) any later version. The GNU C Library is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU Library General Public License for more details. You should have received a copy of the GNU Library General Public License along with the GNU C Library; see the file COPYING.LIB. If not, write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA. */ 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; }