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Tom Rini10e47792018-05-06 17:58:06 -04001// SPDX-License-Identifier: GPL-2.0+
wdenk4a5b6a32001-04-28 17:59:11 +00002/* Copyright (C) 1992, 1997 Free Software Foundation, Inc.
3 This file is part of the GNU C Library.
Wolfgang Denkf3d4e692013-07-08 11:58:49 +02004 */
wdenk4a5b6a32001-04-28 17:59:11 +00005
6typedef struct {
wdenk57b2d802003-06-27 21:31:46 +00007 long quot;
8 long rem;
wdenk4a5b6a32001-04-28 17:59:11 +00009} ldiv_t;
10/* Return the `ldiv_t' representation of NUMER over DENOM. */
11ldiv_t
12ldiv (long int numer, long int denom)
13{
14 ldiv_t result;
15
16 result.quot = numer / denom;
17 result.rem = numer % denom;
18
19 /* The ANSI standard says that |QUOT| <= |NUMER / DENOM|, where
20 NUMER / DENOM is to be computed in infinite precision. In
21 other words, we should always truncate the quotient towards
22 zero, never -infinity. Machine division and remainer may
23 work either way when one or both of NUMER or DENOM is
24 negative. If only one is negative and QUOT has been
25 truncated towards -infinity, REM will have the same sign as
26 DENOM and the opposite sign of NUMER; if both are negative
27 and QUOT has been truncated towards -infinity, REM will be
28 positive (will have the opposite sign of NUMER). These are
29 considered `wrong'. If both are NUM and DENOM are positive,
30 RESULT will always be positive. This all boils down to: if
31 NUMER >= 0, but REM < 0, we got the wrong answer. In that
32 case, to get the right answer, add 1 to QUOT and subtract
33 DENOM from REM. */
34
35 if (numer >= 0 && result.rem < 0)
36 {
37 ++result.quot;
38 result.rem -= denom;
39 }
40
41 return result;
42}