upstream u-boot with additional patches for our devices/boards:
https://lists.denx.de/pipermail/u-boot/2017-March/282789.html (AXP crashes) ;
Gbit ethernet patch for some LIME2 revisions ;
with SPI flash support
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42 lines
1.4 KiB
42 lines
1.4 KiB
// 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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