ldiv.c 1.39 KB
 wdenk committed Apr 28, 2001 1 2 3 ``````/* Copyright (C) 1992, 1997 Free Software Foundation, Inc. This file is part of the GNU C Library. `````` Wolfgang Denk committed Jul 24, 2013 4 5 `````` * SPDX-License-Identifier: LGPL-2.0+ */ `````` wdenk committed Apr 28, 2001 6 7 `````` typedef struct { `````` wdenk committed Jun 27, 2003 8 9 `````` long quot; long rem; `````` wdenk committed Apr 28, 2001 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 ``````} 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; }``````