1 /* Copyright (C) 1992, 1997 Free Software Foundation, Inc. 2 This file is part of the GNU C Library. 3 4 The GNU C Library is free software; you can redistribute it and/or 5 modify it under the terms of the GNU Library General Public License as 6 published by the Free Software Foundation; either version 2 of the 7 License, or (at your option) any later version. 8 9 The GNU C Library is distributed in the hope that it will be useful, 10 but WITHOUT ANY WARRANTY; without even the implied warranty of 11 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU 12 Library General Public License for more details. 13 14 You should have received a copy of the GNU Library General Public 15 License along with the GNU C Library; see the file COPYING.LIB. If not, 16 write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330, 17 Boston, MA 02111-1307, USA. */ 18 19 typedef struct { 20 long quot; 21 long rem; 22 } ldiv_t; 23 /* Return the `ldiv_t' representation of NUMER over DENOM. */ 24 ldiv_t 25 ldiv (long int numer, long int denom) 26 { 27 ldiv_t result; 28 29 result.quot = numer / denom; 30 result.rem = numer % denom; 31 32 /* The ANSI standard says that |QUOT| <= |NUMER / DENOM|, where 33 NUMER / DENOM is to be computed in infinite precision. In 34 other words, we should always truncate the quotient towards 35 zero, never -infinity. Machine division and remainer may 36 work either way when one or both of NUMER or DENOM is 37 negative. If only one is negative and QUOT has been 38 truncated towards -infinity, REM will have the same sign as 39 DENOM and the opposite sign of NUMER; if both are negative 40 and QUOT has been truncated towards -infinity, REM will be 41 positive (will have the opposite sign of NUMER). These are 42 considered `wrong'. If both are NUM and DENOM are positive, 43 RESULT will always be positive. This all boils down to: if 44 NUMER >= 0, but REM < 0, we got the wrong answer. In that 45 case, to get the right answer, add 1 to QUOT and subtract 46 DENOM from REM. */ 47 48 if (numer >= 0 && result.rem < 0) 49 { 50 ++result.quot; 51 result.rem -= denom; 52 } 53 54 return result; 55 } 56