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Import various libgcc functions from syslinux.

Experimentation reveals that gcc ignores -mrtd for the implicit
arithmetic functions (e.g. __udivdi3), but not for the implicit
memcpy() and memset() functions.  Mark the implicit arithmetic
functions with __attribute__((cdecl)) to compensate for this.

(Note: we cannot mark with with __cdecl, because we define __cdecl to
incorporate regparm(0) as well.)
tags/v0.9.3
Michael Brown 17 vuotta sitten
vanhempi
commit
4ce8d61a5c

+ 1
- 0
src/Makefile Näytä tiedosto

@@ -145,6 +145,7 @@ DEBUG_TARGETS	+= dbg%.o c s
145 145
 
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 # SRCDIRS lists all directories containing source files.
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 #
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+SRCDIRS		+= libgcc
148 149
 SRCDIRS		+= core
149 150
 SRCDIRS		+= proto
150 151
 SRCDIRS		+= net net/tcp net/udp

+ 0
- 336
src/arch/i386/core/udivmod64.c Näytä tiedosto

@@ -1,336 +0,0 @@
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-/*
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- * Copyright (C) 2007 Michael Brown <mbrown@fensystems.co.uk>.
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- *
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- * This program is free software; you can redistribute it and/or
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- * modify it under the terms of the GNU General Public License as
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- * published by the Free Software Foundation; either version 2 of the
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- * License, or any later version.
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- *
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- * This program is distributed in the hope that it will be useful, but
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- * WITHOUT ANY WARRANTY; without even the implied warranty of
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- * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
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- * General Public License for more details.
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- *
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- * You should have received a copy of the GNU General Public License
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- * along with this program; if not, write to the Free Software
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- * Foundation, Inc., 675 Mass Ave, Cambridge, MA 02139, USA.
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- */
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-
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-/** @file
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- *
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- * 64-bit division
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- *
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- * The x86 CPU (386 upwards) has a divl instruction which will perform
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- * unsigned division of a 64-bit dividend by a 32-bit divisor.  If the
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- * resulting quotient does not fit in 32 bits, then a CPU exception
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- * will occur.
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- *
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- * Unsigned integer division is expressed as solving 
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- *
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- *   x = d.q + r			0 <= q, 0 <= r < d
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- *
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- * given the dividend (x) and divisor (d), to find the quotient (q)
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- * and remainder (r).
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- *
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- * The x86 divl instruction will solve
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- *
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- *   x = d.q + r			0 <= q, 0 <= r < d
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- *
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- * given x in the range 0 <= x < 2^64 and 1 <= d < 2^32, and causing a
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- * hardware exception if the resulting q >= 2^32.
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- *
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- * We can therefore use divl only if we can prove that the conditions
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- *
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- *   0 <= x < 2^64
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- *   1 <= d < 2^32
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- *        q < 2^32
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- *
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- * are satisfied.
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- *
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- *
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- * Case 1 : 1 <= d < 2^32
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- * ======================
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- *
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- * We express x as
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- *
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- *   x = xh.2^32 + xl			0 <= xh < 2^32, 0 <= xl < 2^32	(1)
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- *
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- * i.e. split x into low and high dwords.  We then solve
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- *
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- *   xh = d.qh + r'			0 <= qh, 0 <= r' < d		(2)
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- *
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- * which we can do using a divl instruction since
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- *
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- *   0 <= xh < 2^64			since 0 <= xh < 2^32 from (1)	(3)
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- *
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- * and
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- *
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- *   1 <= d < 2^32			by definition of this Case	(4)
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- *
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- * and
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- *
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- *   d.qh = xh - r'			from (2)
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- *   d.qh <= xh				since r' >= 0 from (2)
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- *   qh <= xh				since d >= 1 from (2)
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- *   qh < 2^32				since xh < 2^32 from (1)	(5)
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- *
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- * Having obtained qh and r', we then solve
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- *
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- *   ( r'.2^32 + xl ) = d.ql + r	0 <= ql, 0 <= r < d		(6)
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- *
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- * which we can do using another divl instruction since
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- *
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- *   xl <= 2^32 - 1			from (1), so
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- *   r'.2^32 + xl <= ( r' + 1 ).2^32 - 1
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- *   r'.2^32 + xl <= d.2^32 - 1		since r' < d from (2)
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- *   r'.2^32 + xl < d.2^32						(7)
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- *   r'.2^32 + xl < 2^64		since d < 2^32 from (4)		(8)
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- *
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- * and
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- *
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- *   1 <= d < 2^32			by definition of this Case	(9)
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- *
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- * and
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- *
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- *   d.ql = ( r'.2^32 + xl ) - r	from (6)
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- *   d.ql <= r'.2^32 + xl		since r >= 0 from (6)
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- *   d.ql < d.2^32			from (7)
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- *   ql < 2^32				since d >= 1 from (2)		(10)
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- *
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- * This then gives us
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- *
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- *   x = xh.2^32 + xl			from (1)
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- *   x = ( d.qh + r' ).2^32 + xl	from (2)
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- *   x = d.qh.2^32 + ( r'.2^32 + xl )
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- *   x = d.qh.2^32 + d.ql + r		from (3)
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- *   x = d.( qh.2^32 + ql ) + r						(11)
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- *
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- * Letting
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- *
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- *   q = qh.2^32 + ql							(12)
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- *
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- * gives
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- *
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- *   x = d.q + r			from (11) and (12)
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- *
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- * which is the solution.
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- *
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- *
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- * This therefore gives us a two-step algorithm:
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- *
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- *   xh = d.qh + r'			0 <= qh, 0 <= r' < d		(2)
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- *   ( r'.2^32 + xl ) = d.ql + r	0 <= ql, 0 <= r < d		(6)
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- *
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- * which translates to
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- *
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- *   %edx:%eax = 0:xh
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- *   divl d
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- *   qh = %eax
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- *   r' = %edx
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- *
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- *   %edx:%eax = r':xl
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- *   divl d
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- *   ql = %eax
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- *   r = %edx
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- *
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- * Note that if
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- *
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- *   xh < d
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- *
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- * (which is a fast dword comparison) then the first divl instruction
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- * can be omitted, since the answer will be
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- *
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- *   qh = 0
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- *   r = xh
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- *
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- *
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- * Case 2 : 2^32 <= d < 2^64
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- * =========================
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- *
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- * We first express d as
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- *
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- *   d = dh.2^k + dl			2^31 <= dh < 2^32,
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- *					0 <= dl < 2^k, 1 <= k <= 32	(1)
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- *
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- * i.e. find the highest bit set in d, subtract 32, and split d into
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- * dh and dl at that point.
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- *
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- * We then express x as
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- *
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- *   x = xh.2^k + xl			0 <= xl < 2^k			(2)
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- *
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- * giving
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- *
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- *   xh.2^k = x - xl			from (2)
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- *   xh.2^k <= x			since xl >= 0 from (1)
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- *   xh.2^k < 2^64			since xh < 2^64 from (1)
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- *   xh < 2^(64-k)							(3)
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- *
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- * We then solve the division
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- *
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- *   xh = dh.q' + r'	            		0 <= r' < dh		(4)
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- *
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- * which we can do using a divl instruction since
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- *
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- *   0 <= xh < 2^64			since x < 2^64 and xh < x
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- *
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- * and
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- *
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- *   1 <= dh < 2^32			from (1)
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- *
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- * and
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- *
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- *   dh.q' = xh - r'			from (4)
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- *   dh.q' <= xh			since r' >= 0 from (4)
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- *   dh.q' < 2^(64-k)			from (3)			(5)
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- *   q'.2^31 <= dh.q'			since dh >= 2^31 from (1)	(6)
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- *   q'.2^31 < 2^(64-k)			from (5) and (6)
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- *   q' < 2^(33-k)
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- *   q' < 2^32				since k >= 1 from (1)		(7)
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- *
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- * This gives us
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- *
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- *   xh.2^k = dh.q'.2^k + r'.2^k	from (4)
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- *   x - xl = ( d - dl ).q' + r'.2^k	from (1) and (2)
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- *   x = d.q' + ( r'.2^k + xl ) - dl.q'					(8)
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- *
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- * Now
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- *
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- *  r'.2^k + xl < r'.2^k + 2^k		since xl < 2^k from (2)
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- *  r'.2^k + xl < ( r' + 1 ).2^k
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- *  r'.2^k + xl < dh.2^k		since r' < dh from (4)
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- *  r'.2^k + xl < ( d - dl )		from (1)			(9)
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- *
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- *
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- * (missing)
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- *
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- *
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- * This gives us two cases to consider:
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- *
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- * case (a):
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- *
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- *   dl.q' <= ( r'.2^k + xl )						(15a)
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- *
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- * in which case
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- *
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- *   x = d.q' + ( r'.2^k + xl - dl.q' )
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- *
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- * is a direct solution to the division, since
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- *
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- *   r'.2^k + xl < d			from (9)
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- *   ( r'.2^k + xl - dl.q' ) < d	since dl >= 0 and q' >= 0
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- *
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- * and
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- *
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- *   0 <= ( r'.2^k + xl - dl.q' )	from (15a)
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- *
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- * case (b):
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- *
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- *   dl.q' > ( r'.2^k + xl )						(15b)
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- *   
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- * Express
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- *
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- *  x = d.(q'-1) + ( r'.2^k + xl ) + ( d - dl.q' )
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- *  
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- *   
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- * (missing)
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- *
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- *
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- * special case: k = 32 cannot be handled with shifts
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- *
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- * (missing)
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- * 
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- */
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-
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-#include <stdint.h>
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-#include <assert.h>
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-
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-typedef uint64_t UDItype;
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-
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-struct uint64 {
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-	uint32_t l;
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-	uint32_t h;
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-};
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-
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-static inline void udivmod64_lo ( const struct uint64 *x,
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-				  const struct uint64 *d,
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-				  struct uint64 *q,
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-				  struct uint64 *r ) {
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-	uint32_t r_dash;
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-
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-	q->h = 0;
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-	r->h = 0;
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-	r_dash = x->h;
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-
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-	if ( x->h >= d->l ) {
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-		__asm__ ( "divl %2"
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-			  : "=&a" ( q->h ), "=&d" ( r_dash )
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-			  : "g" ( d->l ), "0" ( x->h ), "1" ( 0 ) );
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-	}
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-
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-	__asm__ ( "divl %2"
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-		  : "=&a" ( q->l ), "=&d" ( r->l )
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-		  : "g" ( d->l ), "0" ( x->l ), "1" ( r_dash ) );
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-}
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-
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-void udivmod64 ( const struct uint64 *x,
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-			const struct uint64 *d,
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-			struct uint64 *q,
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-			struct uint64 *r ) {
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-
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-	if ( d->h == 0 ) {
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-		udivmod64_lo ( x, d, q, r );
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-	} else {
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-		assert ( 0 );
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-		while ( 1 ) {};
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-	}	
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-}
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-
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-/**
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- * 64-bit division with remainder
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- *
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- * @v x			Dividend
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- * @v d			Divisor
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- * @ret r		Remainder
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- * @ret q		Quotient
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- */
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-UDItype __udivmoddi4 ( UDItype x, UDItype d, UDItype *r ) {
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-	UDItype q;
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-	UDItype *_x = &x;
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-	UDItype *_d = &d;
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-	UDItype *_q = &q;
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-	UDItype *_r = r;
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-
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-	udivmod64 ( ( struct uint64 * ) _x, ( struct uint64 * ) _d,
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-		    ( struct uint64 * ) _q, ( struct uint64 * ) _r );
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-
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-	assert ( ( x == ( ( d * q ) + (*r) ) ) );
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-	assert ( (*r) < d );
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-
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-	return q;
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-}
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-
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-/**
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- * 64-bit division
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- *
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- * @v x			Dividend
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- * @v d			Divisor
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- * @ret q		Quotient
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- */
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-UDItype __udivdi3 ( UDItype x, UDItype d ) {
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-	UDItype r;
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-	return __udivmoddi4 ( x, d, &r );
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-}
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-
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-/**
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- * 64-bit modulus
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- *
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- * @v x			Dividend
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- * @v d			Divisor
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- * @ret q		Quotient
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- */
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-UDItype __umoddi3 ( UDItype x, UDItype d ) {
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-	UDItype r;
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-	__udivmoddi4 ( x, d, &r );
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-	return r;
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-}

+ 26
- 0
src/libgcc/__divdi3.c Näytä tiedosto

@@ -0,0 +1,26 @@
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+/*
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+ * arch/i386/libgcc/__divdi3.c
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+ */
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+
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+#include "libgcc.h"
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+
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+LIBGCC int64_t __divdi3(int64_t num, int64_t den)
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+{
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+  int minus = 0;
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+  int64_t v;
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+
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+  if ( num < 0 ) {
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+    num = -num;
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+    minus = 1;
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+  }
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+  if ( den < 0 ) {
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+    den = -den;
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+    minus ^= 1;
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+  }
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+
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+  v = __udivmoddi4(num, den, NULL);
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+  if ( minus )
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+    v = -v;
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+
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+  return v;
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+}

+ 26
- 0
src/libgcc/__moddi3.c Näytä tiedosto

@@ -0,0 +1,26 @@
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+/*
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+ * arch/i386/libgcc/__moddi3.c
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+ */
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+
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+#include "libgcc.h"
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+
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+LIBGCC int64_t __moddi3(int64_t num, int64_t den)
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+{
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+  int minus = 0;
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+  int64_t v;
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+
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+  if ( num < 0 ) {
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+    num = -num;
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+    minus = 1;
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+  }
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+  if ( den < 0 ) {
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+    den = -den;
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+    minus ^= 1;
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+  }
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+
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+  (void) __udivmoddi4(num, den, (uint64_t *)&v);
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+  if ( minus )
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+    v = -v;
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+
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+  return v;
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+}

+ 10
- 0
src/libgcc/__udivdi3.c Näytä tiedosto

@@ -0,0 +1,10 @@
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+/*
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+ * arch/i386/libgcc/__divdi3.c
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+ */
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+
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+#include "libgcc.h"
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+
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+LIBGCC uint64_t __udivdi3(uint64_t num, uint64_t den)
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+{
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+  return __udivmoddi4(num, den, NULL);
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+}

+ 32
- 0
src/libgcc/__udivmoddi4.c Näytä tiedosto

@@ -0,0 +1,32 @@
1
+#include "libgcc.h"
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+
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+LIBGCC uint64_t __udivmoddi4(uint64_t num, uint64_t den, uint64_t *rem_p)
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+{
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+  uint64_t quot = 0, qbit = 1;
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+
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+  if ( den == 0 ) {
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+    return 1/((unsigned)den); /* Intentional divide by zero, without
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+				 triggering a compiler warning which
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+				 would abort the build */
11
+  }
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+
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+  /* Left-justify denominator and count shift */
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+  while ( (int64_t)den >= 0 ) {
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+    den <<= 1;
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+    qbit <<= 1;
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+  }
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+
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+  while ( qbit ) {
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+    if ( den <= num ) {
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+      num -= den;
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+      quot += qbit;
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+    }
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+    den >>= 1;
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+    qbit >>= 1;
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+  }
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+
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+  if ( rem_p )
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+    *rem_p = num;
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+
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+  return quot;
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+}

+ 13
- 0
src/libgcc/__umoddi3.c Näytä tiedosto

@@ -0,0 +1,13 @@
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+/*
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+ * arch/i386/libgcc/__umoddi3.c
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+ */
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+
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+#include "libgcc.h"
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+
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+LIBGCC uint64_t __umoddi3(uint64_t num, uint64_t den)
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+{
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+  uint64_t v;
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+
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+  (void) __udivmoddi4(num, den, &v);
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+  return v;
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+}

+ 26
- 0
src/libgcc/libgcc.h Näytä tiedosto

@@ -0,0 +1,26 @@
1
+#ifndef _LIBGCC_H
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+#define _LIBGCC_H
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+
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+#include <stdint.h>
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+#include <stddef.h>
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+
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+/*
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+ * It seems as though gcc expects its implicit arithmetic functions to
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+ * be cdecl, even if -mrtd is specified.  This is somewhat
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+ * inconsistent; for example, if -mregparm=3 is used then the implicit
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+ * functions do become regparm(3).
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+ *
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+ * The implicit calls to memcpy() and memset() which gcc can generate
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+ * do not seem to have this inconsistency; -mregparm and -mrtd affect
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+ * them in the same way as any other function.
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+ *
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+ */
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+#define LIBGCC __attribute__ (( cdecl ))
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+
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+extern LIBGCC uint64_t __udivmoddi4(uint64_t num, uint64_t den, uint64_t *rem);
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+extern LIBGCC uint64_t __udivdi3(uint64_t num, uint64_t den);
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+extern LIBGCC uint64_t __umoddi3(uint64_t num, uint64_t den);
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+extern LIBGCC int64_t __divdi3(int64_t num, int64_t den);
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+extern LIBGCC int64_t __moddi3(int64_t num, int64_t den);
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+
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+#endif /* _LIBGCC_H */

src/core/gcc_implicit.c → src/libgcc/memcpy.c Näytä tiedosto

@@ -1,6 +1,4 @@
1 1
 /** @file
2
- *
3
- * gcc implicit functions
4 2
  *
5 3
  * gcc sometimes likes to insert implicit calls to memcpy().
6 4
  * Unfortunately, there doesn't seem to be any way to prevent it from

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