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 ```1 2 3 4 5 6 7 8 9 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 ``` ``````/* 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; } ``````