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.)

src/Makefile

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

src/arch/i386/core/udivmod64.c

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

src/libgcc/__divdi3.c

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

src/libgcc/__moddi3.c

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

src/libgcc/__udivdi3.c

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

src/libgcc/__udivmoddi4.c

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

src/libgcc/__umoddi3.c

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

src/libgcc/libgcc.h

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

src/libgcc/memcpy.c

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