u-boot/lib/ldiv.c
Tom Rini 83d290c56f SPDX: Convert all of our single license tags to Linux Kernel style
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>
2018-05-07 09:34:12 -04:00

43 lines
1.4 KiB
C

// SPDX-License-Identifier: GPL-2.0+
/* Copyright (C) 1992, 1997 Free Software Foundation, Inc.
This file is part of the GNU C Library.
*/
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;
}