1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
|
.file "div_Xsig.S"
/*---------------------------------------------------------------------------+
| div_Xsig.S |
| |
| Division subroutine for 96 bit quantities |
| |
| Copyright (C) 1994,1995 |
| W. Metzenthen, 22 Parker St, Ormond, Vic 3163, |
| Australia. E-mail billm@jacobi.maths.monash.edu.au |
| |
| |
+---------------------------------------------------------------------------*/
/*---------------------------------------------------------------------------+
| Divide the 96 bit quantity pointed to by a, by that pointed to by b, and |
| put the 96 bit result at the location d. |
| |
| The result may not be accurate to 96 bits. It is intended for use where |
| a result better than 64 bits is required. The result should usually be |
| good to at least 94 bits. |
| The returned result is actually divided by one half. This is done to |
| prevent overflow. |
| |
| .aaaaaaaaaaaaaa / .bbbbbbbbbbbbb -> .dddddddddddd |
| |
| void div_Xsig(Xsig *a, Xsig *b, Xsig *dest) |
| |
+---------------------------------------------------------------------------*/
#include "exception.h"
#include "fpu_emu.h"
#define XsigLL(x) (x)
#define XsigL(x) 4(x)
#define XsigH(x) 8(x)
#ifndef NON_REENTRANT_FPU
/*
Local storage on the stack:
Accumulator: FPU_accum_3:FPU_accum_2:FPU_accum_1:FPU_accum_0
*/
#define FPU_accum_3 -4(%ebp)
#define FPU_accum_2 -8(%ebp)
#define FPU_accum_1 -12(%ebp)
#define FPU_accum_0 -16(%ebp)
#define FPU_result_3 -20(%ebp)
#define FPU_result_2 -24(%ebp)
#define FPU_result_1 -28(%ebp)
#else
.data
/*
Local storage in a static area:
Accumulator: FPU_accum_3:FPU_accum_2:FPU_accum_1:FPU_accum_0
*/
.align 4,0
FPU_accum_3:
.long 0
FPU_accum_2:
.long 0
FPU_accum_1:
.long 0
FPU_accum_0:
.long 0
FPU_result_3:
.long 0
FPU_result_2:
.long 0
FPU_result_1:
.long 0
#endif /* NON_REENTRANT_FPU */
.text
ENTRY(div_Xsig)
pushl %ebp
movl %esp,%ebp
#ifndef NON_REENTRANT_FPU
subl $28,%esp
#endif /* NON_REENTRANT_FPU */
pushl %esi
pushl %edi
pushl %ebx
movl PARAM1,%esi /* pointer to num */
movl PARAM2,%ebx /* pointer to denom */
#ifdef PARANOID
testl $0x80000000, XsigH(%ebx) /* Divisor */
je L_bugged
#endif /* PARANOID */
/*---------------------------------------------------------------------------+
| Divide: Return arg1/arg2 to arg3. |
| |
| The maximum returned value is (ignoring exponents) |
| .ffffffff ffffffff |
| ------------------ = 1.ffffffff fffffffe |
| .80000000 00000000 |
| and the minimum is |
| .80000000 00000000 |
| ------------------ = .80000000 00000001 (rounded) |
| .ffffffff ffffffff |
| |
+---------------------------------------------------------------------------*/
/* Save extended dividend in local register */
/* Divide by 2 to prevent overflow */
clc
movl XsigH(%esi),%eax
rcrl %eax
movl %eax,FPU_accum_3
movl XsigL(%esi),%eax
rcrl %eax
movl %eax,FPU_accum_2
movl XsigLL(%esi),%eax
rcrl %eax
movl %eax,FPU_accum_1
movl $0,%eax
rcrl %eax
movl %eax,FPU_accum_0
movl FPU_accum_2,%eax /* Get the current num */
movl FPU_accum_3,%edx
/*----------------------------------------------------------------------*/
/* Initialization done.
Do the first 32 bits. */
/* We will divide by a number which is too large */
movl XsigH(%ebx),%ecx
addl $1,%ecx
jnc LFirst_div_not_1
/* here we need to divide by 100000000h,
i.e., no division at all.. */
mov %edx,%eax
jmp LFirst_div_done
LFirst_div_not_1:
divl %ecx /* Divide the numerator by the augmented
denom ms dw */
LFirst_div_done:
movl %eax,FPU_result_3 /* Put the result in the answer */
mull XsigH(%ebx) /* mul by the ms dw of the denom */
subl %eax,FPU_accum_2 /* Subtract from the num local reg */
sbbl %edx,FPU_accum_3
movl FPU_result_3,%eax /* Get the result back */
mull XsigL(%ebx) /* now mul the ls dw of the denom */
subl %eax,FPU_accum_1 /* Subtract from the num local reg */
sbbl %edx,FPU_accum_2
sbbl $0,FPU_accum_3
je LDo_2nd_32_bits /* Must check for non-zero result here */
#ifdef PARANOID
jb L_bugged_1
#endif /* PARANOID */
/* need to subtract another once of the denom */
incl FPU_result_3 /* Correct the answer */
movl XsigL(%ebx),%eax
movl XsigH(%ebx),%edx
subl %eax,FPU_accum_1 /* Subtract from the num local reg */
sbbl %edx,FPU_accum_2
#ifdef PARANOID
sbbl $0,FPU_accum_3
jne L_bugged_1 /* Must check for non-zero result here */
#endif /* PARANOID */
/*----------------------------------------------------------------------*/
/* Half of the main problem is done, there is just a reduced numerator
to handle now.
Work with the second 32 bits, FPU_accum_0 not used from now on */
LDo_2nd_32_bits:
movl FPU_accum_2,%edx /* get the reduced num */
movl FPU_accum_1,%eax
/* need to check for possible subsequent overflow */
cmpl XsigH(%ebx),%edx
jb LDo_2nd_div
ja LPrevent_2nd_overflow
cmpl XsigL(%ebx),%eax
jb LDo_2nd_div
LPrevent_2nd_overflow:
/* The numerator is greater or equal, would cause overflow */
/* prevent overflow */
subl XsigL(%ebx),%eax
sbbl XsigH(%ebx),%edx
movl %edx,FPU_accum_2
movl %eax,FPU_accum_1
incl FPU_result_3 /* Reflect the subtraction in the answer */
#ifdef PARANOID
je L_bugged_2 /* Can't bump the result to 1.0 */
#endif /* PARANOID */
LDo_2nd_div:
cmpl $0,%ecx /* augmented denom msw */
jnz LSecond_div_not_1
/* %ecx == 0, we are dividing by 1.0 */
mov %edx,%eax
jmp LSecond_div_done
LSecond_div_not_1:
divl %ecx /* Divide the numerator by the denom ms dw */
LSecond_div_done:
movl %eax,FPU_result_2 /* Put the result in the answer */
mull XsigH(%ebx) /* mul by the ms dw of the denom */
subl %eax,FPU_accum_1 /* Subtract from the num local reg */
sbbl %edx,FPU_accum_2
#ifdef PARANOID
jc L_bugged_2
#endif /* PARANOID */
movl FPU_result_2,%eax /* Get the result back */
mull XsigL(%ebx) /* now mul the ls dw of the denom */
subl %eax,FPU_accum_0 /* Subtract from the num local reg */
sbbl %edx,FPU_accum_1 /* Subtract from the num local reg */
sbbl $0,FPU_accum_2
#ifdef PARANOID
jc L_bugged_2
#endif /* PARANOID */
jz LDo_3rd_32_bits
#ifdef PARANOID
cmpl $1,FPU_accum_2
jne L_bugged_2
#endif /* PARANOID */
/* need to subtract another once of the denom */
movl XsigL(%ebx),%eax
movl XsigH(%ebx),%edx
subl %eax,FPU_accum_0 /* Subtract from the num local reg */
sbbl %edx,FPU_accum_1
sbbl $0,FPU_accum_2
#ifdef PARANOID
jc L_bugged_2
jne L_bugged_2
#endif /* PARANOID */
addl $1,FPU_result_2 /* Correct the answer */
adcl $0,FPU_result_3
#ifdef PARANOID
jc L_bugged_2 /* Must check for non-zero result here */
#endif /* PARANOID */
/*----------------------------------------------------------------------*/
/* The division is essentially finished here, we just need to perform
tidying operations.
Deal with the 3rd 32 bits */
LDo_3rd_32_bits:
/* We use an approximation for the third 32 bits.
To take account of the 3rd 32 bits of the divisor
(call them del), we subtract del * (a/b) */
movl FPU_result_3,%eax /* a/b */
mull XsigLL(%ebx) /* del */
subl %edx,FPU_accum_1
/* A borrow indicates that the result is negative */
jnb LTest_over
movl XsigH(%ebx),%edx
addl %edx,FPU_accum_1
subl $1,FPU_result_2 /* Adjust the answer */
sbbl $0,FPU_result_3
/* The above addition might not have been enough, check again. */
movl FPU_accum_1,%edx /* get the reduced num */
cmpl XsigH(%ebx),%edx /* denom */
jb LDo_3rd_div
movl XsigH(%ebx),%edx
addl %edx,FPU_accum_1
subl $1,FPU_result_2 /* Adjust the answer */
sbbl $0,FPU_result_3
jmp LDo_3rd_div
LTest_over:
movl FPU_accum_1,%edx /* get the reduced num */
/* need to check for possible subsequent overflow */
cmpl XsigH(%ebx),%edx /* denom */
jb LDo_3rd_div
/* prevent overflow */
subl XsigH(%ebx),%edx
movl %edx,FPU_accum_1
addl $1,FPU_result_2 /* Reflect the subtraction in the answer */
adcl $0,FPU_result_3
LDo_3rd_div:
movl FPU_accum_0,%eax
movl FPU_accum_1,%edx
divl XsigH(%ebx)
movl %eax,FPU_result_1 /* Rough estimate of third word */
movl PARAM3,%esi /* pointer to answer */
movl FPU_result_1,%eax
movl %eax,XsigLL(%esi)
movl FPU_result_2,%eax
movl %eax,XsigL(%esi)
movl FPU_result_3,%eax
movl %eax,XsigH(%esi)
L_exit:
popl %ebx
popl %edi
popl %esi
leave
ret
#ifdef PARANOID
/* The logic is wrong if we got here */
L_bugged:
pushl EX_INTERNAL|0x240
call EXCEPTION
pop %ebx
jmp L_exit
L_bugged_1:
pushl EX_INTERNAL|0x241
call EXCEPTION
pop %ebx
jmp L_exit
L_bugged_2:
pushl EX_INTERNAL|0x242
call EXCEPTION
pop %ebx
jmp L_exit
#endif /* PARANOID */
ENDPROC(div_Xsig)
|