Переглянути джерело

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 роки тому
джерело
коміт
4ce8d61a5c

+ 1
- 0
src/Makefile Переглянути файл

@@ -145,6 +145,7 @@ DEBUG_TARGETS	+= dbg%.o c s
145 145
 
146 146
 # SRCDIRS lists all directories containing source files.
147 147
 #
148
+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 Переглянути файл

@@ -1,336 +0,0 @@
1
-/*
2
- * Copyright (C) 2007 Michael Brown <mbrown@fensystems.co.uk>.
3
- *
4
- * This program is free software; you can redistribute it and/or
5
- * modify it under the terms of the GNU General Public License as
6
- * published by the Free Software Foundation; either version 2 of the
7
- * License, or any later version.
8
- *
9
- * This program is distributed in the hope that it will be useful, but
10
- * WITHOUT ANY WARRANTY; without even the implied warranty of
11
- * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
12
- * General Public License for more details.
13
- *
14
- * You should have received a copy of the GNU General Public License
15
- * along with this program; if not, write to the Free Software
16
- * Foundation, Inc., 675 Mass Ave, Cambridge, MA 02139, USA.
17
- */
18
-
19
-/** @file
20
- *
21
- * 64-bit division
22
- *
23
- * The x86 CPU (386 upwards) has a divl instruction which will perform
24
- * unsigned division of a 64-bit dividend by a 32-bit divisor.  If the
25
- * resulting quotient does not fit in 32 bits, then a CPU exception
26
- * will occur.
27
- *
28
- * Unsigned integer division is expressed as solving 
29
- *
30
- *   x = d.q + r			0 <= q, 0 <= r < d
31
- *
32
- * given the dividend (x) and divisor (d), to find the quotient (q)
33
- * and remainder (r).
34
- *
35
- * The x86 divl instruction will solve
36
- *
37
- *   x = d.q + r			0 <= q, 0 <= r < d
38
- *
39
- * given x in the range 0 <= x < 2^64 and 1 <= d < 2^32, and causing a
40
- * hardware exception if the resulting q >= 2^32.
41
- *
42
- * We can therefore use divl only if we can prove that the conditions
43
- *
44
- *   0 <= x < 2^64
45
- *   1 <= d < 2^32
46
- *        q < 2^32
47
- *
48
- * are satisfied.
49
- *
50
- *
51
- * Case 1 : 1 <= d < 2^32
52
- * ======================
53
- *
54
- * We express x as
55
- *
56
- *   x = xh.2^32 + xl			0 <= xh < 2^32, 0 <= xl < 2^32	(1)
57
- *
58
- * i.e. split x into low and high dwords.  We then solve
59
- *
60
- *   xh = d.qh + r'			0 <= qh, 0 <= r' < d		(2)
61
- *
62
- * which we can do using a divl instruction since
63
- *
64
- *   0 <= xh < 2^64			since 0 <= xh < 2^32 from (1)	(3)
65
- *
66
- * and
67
- *
68
- *   1 <= d < 2^32			by definition of this Case	(4)
69
- *
70
- * and
71
- *
72
- *   d.qh = xh - r'			from (2)
73
- *   d.qh <= xh				since r' >= 0 from (2)
74
- *   qh <= xh				since d >= 1 from (2)
75
- *   qh < 2^32				since xh < 2^32 from (1)	(5)
76
- *
77
- * Having obtained qh and r', we then solve
78
- *
79
- *   ( r'.2^32 + xl ) = d.ql + r	0 <= ql, 0 <= r < d		(6)
80
- *
81
- * which we can do using another divl instruction since
82
- *
83
- *   xl <= 2^32 - 1			from (1), so
84
- *   r'.2^32 + xl <= ( r' + 1 ).2^32 - 1
85
- *   r'.2^32 + xl <= d.2^32 - 1		since r' < d from (2)
86
- *   r'.2^32 + xl < d.2^32						(7)
87
- *   r'.2^32 + xl < 2^64		since d < 2^32 from (4)		(8)
88
- *
89
- * and
90
- *
91
- *   1 <= d < 2^32			by definition of this Case	(9)
92
- *
93
- * and
94
- *
95
- *   d.ql = ( r'.2^32 + xl ) - r	from (6)
96
- *   d.ql <= r'.2^32 + xl		since r >= 0 from (6)
97
- *   d.ql < d.2^32			from (7)
98
- *   ql < 2^32				since d >= 1 from (2)		(10)
99
- *
100
- * This then gives us
101
- *
102
- *   x = xh.2^32 + xl			from (1)
103
- *   x = ( d.qh + r' ).2^32 + xl	from (2)
104
- *   x = d.qh.2^32 + ( r'.2^32 + xl )
105
- *   x = d.qh.2^32 + d.ql + r		from (3)
106
- *   x = d.( qh.2^32 + ql ) + r						(11)
107
- *
108
- * Letting
109
- *
110
- *   q = qh.2^32 + ql							(12)
111
- *
112
- * gives
113
- *
114
- *   x = d.q + r			from (11) and (12)
115
- *
116
- * which is the solution.
117
- *
118
- *
119
- * This therefore gives us a two-step algorithm:
120
- *
121
- *   xh = d.qh + r'			0 <= qh, 0 <= r' < d		(2)
122
- *   ( r'.2^32 + xl ) = d.ql + r	0 <= ql, 0 <= r < d		(6)
123
- *
124
- * which translates to
125
- *
126
- *   %edx:%eax = 0:xh
127
- *   divl d
128
- *   qh = %eax
129
- *   r' = %edx
130
- *
131
- *   %edx:%eax = r':xl
132
- *   divl d
133
- *   ql = %eax
134
- *   r = %edx
135
- *
136
- * Note that if
137
- *
138
- *   xh < d
139
- *
140
- * (which is a fast dword comparison) then the first divl instruction
141
- * can be omitted, since the answer will be
142
- *
143
- *   qh = 0
144
- *   r = xh
145
- *
146
- *
147
- * Case 2 : 2^32 <= d < 2^64
148
- * =========================
149
- *
150
- * We first express d as
151
- *
152
- *   d = dh.2^k + dl			2^31 <= dh < 2^32,
153
- *					0 <= dl < 2^k, 1 <= k <= 32	(1)
154
- *
155
- * i.e. find the highest bit set in d, subtract 32, and split d into
156
- * dh and dl at that point.
157
- *
158
- * We then express x as
159
- *
160
- *   x = xh.2^k + xl			0 <= xl < 2^k			(2)
161
- *
162
- * giving
163
- *
164
- *   xh.2^k = x - xl			from (2)
165
- *   xh.2^k <= x			since xl >= 0 from (1)
166
- *   xh.2^k < 2^64			since xh < 2^64 from (1)
167
- *   xh < 2^(64-k)							(3)
168
- *
169
- * We then solve the division
170
- *
171
- *   xh = dh.q' + r'	            		0 <= r' < dh		(4)
172
- *
173
- * which we can do using a divl instruction since
174
- *
175
- *   0 <= xh < 2^64			since x < 2^64 and xh < x
176
- *
177
- * and
178
- *
179
- *   1 <= dh < 2^32			from (1)
180
- *
181
- * and
182
- *
183
- *   dh.q' = xh - r'			from (4)
184
- *   dh.q' <= xh			since r' >= 0 from (4)
185
- *   dh.q' < 2^(64-k)			from (3)			(5)
186
- *   q'.2^31 <= dh.q'			since dh >= 2^31 from (1)	(6)
187
- *   q'.2^31 < 2^(64-k)			from (5) and (6)
188
- *   q' < 2^(33-k)
189
- *   q' < 2^32				since k >= 1 from (1)		(7)
190
- *
191
- * This gives us
192
- *
193
- *   xh.2^k = dh.q'.2^k + r'.2^k	from (4)
194
- *   x - xl = ( d - dl ).q' + r'.2^k	from (1) and (2)
195
- *   x = d.q' + ( r'.2^k + xl ) - dl.q'					(8)
196
- *
197
- * Now
198
- *
199
- *  r'.2^k + xl < r'.2^k + 2^k		since xl < 2^k from (2)
200
- *  r'.2^k + xl < ( r' + 1 ).2^k
201
- *  r'.2^k + xl < dh.2^k		since r' < dh from (4)
202
- *  r'.2^k + xl < ( d - dl )		from (1)			(9)
203
- *
204
- *
205
- * (missing)
206
- *
207
- *
208
- * This gives us two cases to consider:
209
- *
210
- * case (a):
211
- *
212
- *   dl.q' <= ( r'.2^k + xl )						(15a)
213
- *
214
- * in which case
215
- *
216
- *   x = d.q' + ( r'.2^k + xl - dl.q' )
217
- *
218
- * is a direct solution to the division, since
219
- *
220
- *   r'.2^k + xl < d			from (9)
221
- *   ( r'.2^k + xl - dl.q' ) < d	since dl >= 0 and q' >= 0
222
- *
223
- * and
224
- *
225
- *   0 <= ( r'.2^k + xl - dl.q' )	from (15a)
226
- *
227
- * case (b):
228
- *
229
- *   dl.q' > ( r'.2^k + xl )						(15b)
230
- *   
231
- * Express
232
- *
233
- *  x = d.(q'-1) + ( r'.2^k + xl ) + ( d - dl.q' )
234
- *  
235
- *   
236
- * (missing)
237
- *
238
- *
239
- * special case: k = 32 cannot be handled with shifts
240
- *
241
- * (missing)
242
- * 
243
- */
244
-
245
-#include <stdint.h>
246
-#include <assert.h>
247
-
248
-typedef uint64_t UDItype;
249
-
250
-struct uint64 {
251
-	uint32_t l;
252
-	uint32_t h;
253
-};
254
-
255
-static inline void udivmod64_lo ( const struct uint64 *x,
256
-				  const struct uint64 *d,
257
-				  struct uint64 *q,
258
-				  struct uint64 *r ) {
259
-	uint32_t r_dash;
260
-
261
-	q->h = 0;
262
-	r->h = 0;
263
-	r_dash = x->h;
264
-
265
-	if ( x->h >= d->l ) {
266
-		__asm__ ( "divl %2"
267
-			  : "=&a" ( q->h ), "=&d" ( r_dash )
268
-			  : "g" ( d->l ), "0" ( x->h ), "1" ( 0 ) );
269
-	}
270
-
271
-	__asm__ ( "divl %2"
272
-		  : "=&a" ( q->l ), "=&d" ( r->l )
273
-		  : "g" ( d->l ), "0" ( x->l ), "1" ( r_dash ) );
274
-}
275
-
276
-void udivmod64 ( const struct uint64 *x,
277
-			const struct uint64 *d,
278
-			struct uint64 *q,
279
-			struct uint64 *r ) {
280
-
281
-	if ( d->h == 0 ) {
282
-		udivmod64_lo ( x, d, q, r );
283
-	} else {
284
-		assert ( 0 );
285
-		while ( 1 ) {};
286
-	}	
287
-}
288
-
289
-/**
290
- * 64-bit division with remainder
291
- *
292
- * @v x			Dividend
293
- * @v d			Divisor
294
- * @ret r		Remainder
295
- * @ret q		Quotient
296
- */
297
-UDItype __udivmoddi4 ( UDItype x, UDItype d, UDItype *r ) {
298
-	UDItype q;
299
-	UDItype *_x = &x;
300
-	UDItype *_d = &d;
301
-	UDItype *_q = &q;
302
-	UDItype *_r = r;
303
-
304
-	udivmod64 ( ( struct uint64 * ) _x, ( struct uint64 * ) _d,
305
-		    ( struct uint64 * ) _q, ( struct uint64 * ) _r );
306
-
307
-	assert ( ( x == ( ( d * q ) + (*r) ) ) );
308
-	assert ( (*r) < d );
309
-
310
-	return q;
311
-}
312
-
313
-/**
314
- * 64-bit division
315
- *
316
- * @v x			Dividend
317
- * @v d			Divisor
318
- * @ret q		Quotient
319
- */
320
-UDItype __udivdi3 ( UDItype x, UDItype d ) {
321
-	UDItype r;
322
-	return __udivmoddi4 ( x, d, &r );
323
-}
324
-
325
-/**
326
- * 64-bit modulus
327
- *
328
- * @v x			Dividend
329
- * @v d			Divisor
330
- * @ret q		Quotient
331
- */
332
-UDItype __umoddi3 ( UDItype x, UDItype d ) {
333
-	UDItype r;
334
-	__udivmoddi4 ( x, d, &r );
335
-	return r;
336
-}

+ 26
- 0
src/libgcc/__divdi3.c Переглянути файл

@@ -0,0 +1,26 @@
1
+/*
2
+ * arch/i386/libgcc/__divdi3.c
3
+ */
4
+
5
+#include "libgcc.h"
6
+
7
+LIBGCC int64_t __divdi3(int64_t num, int64_t den)
8
+{
9
+  int minus = 0;
10
+  int64_t v;
11
+
12
+  if ( num < 0 ) {
13
+    num = -num;
14
+    minus = 1;
15
+  }
16
+  if ( den < 0 ) {
17
+    den = -den;
18
+    minus ^= 1;
19
+  }
20
+
21
+  v = __udivmoddi4(num, den, NULL);
22
+  if ( minus )
23
+    v = -v;
24
+
25
+  return v;
26
+}

+ 26
- 0
src/libgcc/__moddi3.c Переглянути файл

@@ -0,0 +1,26 @@
1
+/*
2
+ * arch/i386/libgcc/__moddi3.c
3
+ */
4
+
5
+#include "libgcc.h"
6
+
7
+LIBGCC int64_t __moddi3(int64_t num, int64_t den)
8
+{
9
+  int minus = 0;
10
+  int64_t v;
11
+
12
+  if ( num < 0 ) {
13
+    num = -num;
14
+    minus = 1;
15
+  }
16
+  if ( den < 0 ) {
17
+    den = -den;
18
+    minus ^= 1;
19
+  }
20
+
21
+  (void) __udivmoddi4(num, den, (uint64_t *)&v);
22
+  if ( minus )
23
+    v = -v;
24
+
25
+  return v;
26
+}

+ 10
- 0
src/libgcc/__udivdi3.c Переглянути файл

@@ -0,0 +1,10 @@
1
+/*
2
+ * arch/i386/libgcc/__divdi3.c
3
+ */
4
+
5
+#include "libgcc.h"
6
+
7
+LIBGCC uint64_t __udivdi3(uint64_t num, uint64_t den)
8
+{
9
+  return __udivmoddi4(num, den, NULL);
10
+}

+ 32
- 0
src/libgcc/__udivmoddi4.c Переглянути файл

@@ -0,0 +1,32 @@
1
+#include "libgcc.h"
2
+
3
+LIBGCC uint64_t __udivmoddi4(uint64_t num, uint64_t den, uint64_t *rem_p)
4
+{
5
+  uint64_t quot = 0, qbit = 1;
6
+
7
+  if ( den == 0 ) {
8
+    return 1/((unsigned)den); /* Intentional divide by zero, without
9
+				 triggering a compiler warning which
10
+				 would abort the build */
11
+  }
12
+
13
+  /* Left-justify denominator and count shift */
14
+  while ( (int64_t)den >= 0 ) {
15
+    den <<= 1;
16
+    qbit <<= 1;
17
+  }
18
+
19
+  while ( qbit ) {
20
+    if ( den <= num ) {
21
+      num -= den;
22
+      quot += qbit;
23
+    }
24
+    den >>= 1;
25
+    qbit >>= 1;
26
+  }
27
+
28
+  if ( rem_p )
29
+    *rem_p = num;
30
+
31
+  return quot;
32
+}

+ 13
- 0
src/libgcc/__umoddi3.c Переглянути файл

@@ -0,0 +1,13 @@
1
+/*
2
+ * arch/i386/libgcc/__umoddi3.c
3
+ */
4
+
5
+#include "libgcc.h"
6
+
7
+LIBGCC uint64_t __umoddi3(uint64_t num, uint64_t den)
8
+{
9
+  uint64_t v;
10
+
11
+  (void) __udivmoddi4(num, den, &v);
12
+  return v;
13
+}

+ 26
- 0
src/libgcc/libgcc.h Переглянути файл

@@ -0,0 +1,26 @@
1
+#ifndef _LIBGCC_H
2
+#define _LIBGCC_H
3
+
4
+#include <stdint.h>
5
+#include <stddef.h>
6
+
7
+/*
8
+ * It seems as though gcc expects its implicit arithmetic functions to
9
+ * be cdecl, even if -mrtd is specified.  This is somewhat
10
+ * inconsistent; for example, if -mregparm=3 is used then the implicit
11
+ * functions do become regparm(3).
12
+ *
13
+ * The implicit calls to memcpy() and memset() which gcc can generate
14
+ * do not seem to have this inconsistency; -mregparm and -mrtd affect
15
+ * them in the same way as any other function.
16
+ *
17
+ */
18
+#define LIBGCC __attribute__ (( cdecl ))
19
+
20
+extern LIBGCC uint64_t __udivmoddi4(uint64_t num, uint64_t den, uint64_t *rem);
21
+extern LIBGCC uint64_t __udivdi3(uint64_t num, uint64_t den);
22
+extern LIBGCC uint64_t __umoddi3(uint64_t num, uint64_t den);
23
+extern LIBGCC int64_t __divdi3(int64_t num, int64_t den);
24
+extern LIBGCC int64_t __moddi3(int64_t num, int64_t den);
25
+
26
+#endif /* _LIBGCC_H */

src/core/gcc_implicit.c → src/libgcc/memcpy.c Переглянути файл

@@ -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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