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1 | /* Copyright (C) 1992, 1997 Free Software Foundation, Inc. |
2 | This file is part of the GNU C Library. | |
3 | ||
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4 | * SPDX-License-Identifier: LGPL-2.0+ |
5 | */ | |
4a5b6a35 WD |
6 | |
7 | typedef struct { | |
8bde7f77 WD |
8 | long quot; |
9 | long rem; | |
4a5b6a35 WD |
10 | } ldiv_t; |
11 | /* Return the `ldiv_t' representation of NUMER over DENOM. */ | |
12 | ldiv_t | |
13 | ldiv (long int numer, long int denom) | |
14 | { | |
15 | ldiv_t result; | |
16 | ||
17 | result.quot = numer / denom; | |
18 | result.rem = numer % denom; | |
19 | ||
20 | /* The ANSI standard says that |QUOT| <= |NUMER / DENOM|, where | |
21 | NUMER / DENOM is to be computed in infinite precision. In | |
22 | other words, we should always truncate the quotient towards | |
23 | zero, never -infinity. Machine division and remainer may | |
24 | work either way when one or both of NUMER or DENOM is | |
25 | negative. If only one is negative and QUOT has been | |
26 | truncated towards -infinity, REM will have the same sign as | |
27 | DENOM and the opposite sign of NUMER; if both are negative | |
28 | and QUOT has been truncated towards -infinity, REM will be | |
29 | positive (will have the opposite sign of NUMER). These are | |
30 | considered `wrong'. If both are NUM and DENOM are positive, | |
31 | RESULT will always be positive. This all boils down to: if | |
32 | NUMER >= 0, but REM < 0, we got the wrong answer. In that | |
33 | case, to get the right answer, add 1 to QUOT and subtract | |
34 | DENOM from REM. */ | |
35 | ||
36 | if (numer >= 0 && result.rem < 0) | |
37 | { | |
38 | ++result.quot; | |
39 | result.rem -= denom; | |
40 | } | |
41 | ||
42 | return result; | |
43 | } |