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https://github.com/brain-hackers/u-boot-brain
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83d290c56f
When U-Boot started using SPDX tags we were among the early adopters and there weren't a lot of other examples to borrow from. So we picked the area of the file that usually had a full license text and replaced it with an appropriate SPDX-License-Identifier: entry. Since then, the Linux Kernel has adopted SPDX tags and they place it as the very first line in a file (except where shebangs are used, then it's second line) and with slightly different comment styles than us. In part due to community overlap, in part due to better tag visibility and in part for other minor reasons, switch over to that style. This commit changes all instances where we have a single declared license in the tag as both the before and after are identical in tag contents. There's also a few places where I found we did not have a tag and have introduced one. Signed-off-by: Tom Rini <trini@konsulko.com>
43 lines
1.4 KiB
C
43 lines
1.4 KiB
C
// SPDX-License-Identifier: GPL-2.0+
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/* Copyright (C) 1992, 1997 Free Software Foundation, Inc.
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This file is part of the GNU C Library.
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*/
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typedef struct {
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long quot;
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long rem;
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} ldiv_t;
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/* Return the `ldiv_t' representation of NUMER over DENOM. */
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ldiv_t
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ldiv (long int numer, long int denom)
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{
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ldiv_t result;
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result.quot = numer / denom;
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result.rem = numer % denom;
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/* The ANSI standard says that |QUOT| <= |NUMER / DENOM|, where
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NUMER / DENOM is to be computed in infinite precision. In
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other words, we should always truncate the quotient towards
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zero, never -infinity. Machine division and remainer may
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work either way when one or both of NUMER or DENOM is
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negative. If only one is negative and QUOT has been
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truncated towards -infinity, REM will have the same sign as
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DENOM and the opposite sign of NUMER; if both are negative
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and QUOT has been truncated towards -infinity, REM will be
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positive (will have the opposite sign of NUMER). These are
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considered `wrong'. If both are NUM and DENOM are positive,
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RESULT will always be positive. This all boils down to: if
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NUMER >= 0, but REM < 0, we got the wrong answer. In that
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case, to get the right answer, add 1 to QUOT and subtract
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DENOM from REM. */
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if (numer >= 0 && result.rem < 0)
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{
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++result.quot;
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result.rem -= denom;
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}
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return result;
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}
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