summary refs log tree commit diff
path: root/sysdeps/sparc/sparc32/rem.S
blob: effacee0d015b574333b7c529bf6a01941e7057b (plain) (blame)
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
   /* This file is generated from divrem.m4; DO NOT EDIT! */
/*
 * Division and remainder, from Appendix E of the Sparc Version 8
 * Architecture Manual, with fixes from Gordon Irlam.
 */

/*
 * Input: dividend and divisor in %o0 and %o1 respectively.
 *
 * m4 parameters:
 *  .rem	name of function to generate
 *  rem		rem=div => %o0 / %o1; rem=rem => %o0 % %o1
 *  true		true=true => signed; true=false => unsigned
 *
 * Algorithm parameters:
 *  N		how many bits per iteration we try to get (4)
 *  WORDSIZE	total number of bits (32)
 *
 * Derived constants:
 *  TOPBITS	number of bits in the top decade of a number
 *
 * Important variables:
 *  Q		the partial quotient under development (initially 0)
 *  R		the remainder so far, initially the dividend
 *  ITER	number of main division loop iterations required;
 *		equal to ceil(log2(quotient) / N).  Note that this
 *		is the log base (2^N) of the quotient.
 *  V		the current comparand, initially divisor*2^(ITER*N-1)
 *
 * Cost:
 *  Current estimate for non-large dividend is
 *	ceil(log2(quotient) / N) * (10 + 7N/2) + C
 *  A large dividend is one greater than 2^(31-TOPBITS) and takes a
 *  different path, as the upper bits of the quotient must be developed
 *  one bit at a time.
 */



#include <sysdep.h>
#include <sys/trap.h>

ENTRY(.rem)
	! compute sign of result; if neither is negative, no problem
	orcc	%o1, %o0, %g0	! either negative?
	bge	2f			! no, go do the divide
	mov	%o0, %g3		! sign of remainder matches %o0
	tst	%o1
	bge	1f
	tst	%o0
	! %o1 is definitely negative; %o0 might also be negative
	bge	2f			! if %o0 not negative...
	sub	%g0, %o1, %o1	! in any case, make %o1 nonneg
1:	! %o0 is negative, %o1 is nonnegative
	sub	%g0, %o0, %o0	! make %o0 nonnegative
2:

	! Ready to divide.  Compute size of quotient; scale comparand.
	orcc	%o1, %g0, %o5
	bne	1f
	mov	%o0, %o3

		! Divide by zero trap.  If it returns, return 0 (about as
		! wrong as possible, but that is what SunOS does...).
		ta	ST_DIV0
		retl
		clr	%o0

1:
	cmp	%o3, %o5			! if %o1 exceeds %o0, done
	blu	LOC(got_result)		! (and algorithm fails otherwise)
	clr	%o2
	sethi	%hi(1 << (32 - 4 - 1)), %g1
	cmp	%o3, %g1
	blu	LOC(not_really_big)
	clr	%o4

	! Here the dividend is >= 2**(31-N) or so.  We must be careful here,
	! as our usual N-at-a-shot divide step will cause overflow and havoc.
	! The number of bits in the result here is N*ITER+SC, where SC <= N.
	! Compute ITER in an unorthodox manner: know we need to shift V into
	! the top decade: so do not even bother to compare to R.
	1:
		cmp	%o5, %g1
		bgeu	3f
		mov	1, %g2
		sll	%o5, 4, %o5
		b	1b
		add	%o4, 1, %o4

	! Now compute %g2.
	2:	addcc	%o5, %o5, %o5
		bcc	LOC(not_too_big)
		add	%g2, 1, %g2

		! We get here if the %o1 overflowed while shifting.
		! This means that %o3 has the high-order bit set.
		! Restore %o5 and subtract from %o3.
		sll	%g1, 4, %g1	! high order bit
		srl	%o5, 1, %o5		! rest of %o5
		add	%o5, %g1, %o5
		b	LOC(do_single_div)
		sub	%g2, 1, %g2

	LOC(not_too_big):
	3:	cmp	%o5, %o3
		blu	2b
		nop
		be	LOC(do_single_div)
		nop
	/* NB: these are commented out in the V8-Sparc manual as well */
	/* (I do not understand this) */
	! %o5 > %o3: went too far: back up 1 step
	!	srl	%o5, 1, %o5
	!	dec	%g2
	! do single-bit divide steps
	!
	! We have to be careful here.  We know that %o3 >= %o5, so we can do the
	! first divide step without thinking.  BUT, the others are conditional,
	! and are only done if %o3 >= 0.  Because both %o3 and %o5 may have the high-
	! order bit set in the first step, just falling into the regular
	! division loop will mess up the first time around.
	! So we unroll slightly...
	LOC(do_single_div):
		subcc	%g2, 1, %g2
		bl	LOC(end_regular_divide)
		nop
		sub	%o3, %o5, %o3
		mov	1, %o2
		b	LOC(end_single_divloop)
		nop
	LOC(single_divloop):
		sll	%o2, 1, %o2
		bl	1f
		srl	%o5, 1, %o5
		! %o3 >= 0
		sub	%o3, %o5, %o3
		b	2f
		add	%o2, 1, %o2
	1:	! %o3 < 0
		add	%o3, %o5, %o3
		sub	%o2, 1, %o2
	2:
	LOC(end_single_divloop):
		subcc	%g2, 1, %g2
		bge	LOC(single_divloop)
		tst	%o3
		b,a	LOC(end_regular_divide)

LOC(not_really_big):
1:
	sll	%o5, 4, %o5
	cmp	%o5, %o3
	bleu	1b
	addcc	%o4, 1, %o4
	be	LOC(got_result)
	sub	%o4, 1, %o4

	tst	%o3	! set up for initial iteration
LOC(divloop):
	sll	%o2, 4, %o2
		! depth 1, accumulated bits 0
	bl	LOC(1.16)
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
			! depth 2, accumulated bits 1
	bl	LOC(2.17)
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
			! depth 3, accumulated bits 3
	bl	LOC(3.19)
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
			! depth 4, accumulated bits 7
	bl	LOC(4.23)
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
		b	9f
		add	%o2, (7*2+1), %o2
	
LOC(4.23):
	! remainder is negative
	addcc	%o3,%o5,%o3
		b	9f
		add	%o2, (7*2-1), %o2
	
	
LOC(3.19):
	! remainder is negative
	addcc	%o3,%o5,%o3
			! depth 4, accumulated bits 5
	bl	LOC(4.21)
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
		b	9f
		add	%o2, (5*2+1), %o2
	
LOC(4.21):
	! remainder is negative
	addcc	%o3,%o5,%o3
		b	9f
		add	%o2, (5*2-1), %o2
	
	
	
LOC(2.17):
	! remainder is negative
	addcc	%o3,%o5,%o3
			! depth 3, accumulated bits 1
	bl	LOC(3.17)
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
			! depth 4, accumulated bits 3
	bl	LOC(4.19)
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
		b	9f
		add	%o2, (3*2+1), %o2
	
LOC(4.19):
	! remainder is negative
	addcc	%o3,%o5,%o3
		b	9f
		add	%o2, (3*2-1), %o2
	
	
LOC(3.17):
	! remainder is negative
	addcc	%o3,%o5,%o3
			! depth 4, accumulated bits 1
	bl	LOC(4.17)
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
		b	9f
		add	%o2, (1*2+1), %o2
	
LOC(4.17):
	! remainder is negative
	addcc	%o3,%o5,%o3
		b	9f
		add	%o2, (1*2-1), %o2
	
	
	
	
LOC(1.16):
	! remainder is negative
	addcc	%o3,%o5,%o3
			! depth 2, accumulated bits -1
	bl	LOC(2.15)
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
			! depth 3, accumulated bits -1
	bl	LOC(3.15)
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
			! depth 4, accumulated bits -1
	bl	LOC(4.15)
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
		b	9f
		add	%o2, (-1*2+1), %o2
	
LOC(4.15):
	! remainder is negative
	addcc	%o3,%o5,%o3
		b	9f
		add	%o2, (-1*2-1), %o2
	
	
LOC(3.15):
	! remainder is negative
	addcc	%o3,%o5,%o3
			! depth 4, accumulated bits -3
	bl	LOC(4.13)
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
		b	9f
		add	%o2, (-3*2+1), %o2
	
LOC(4.13):
	! remainder is negative
	addcc	%o3,%o5,%o3
		b	9f
		add	%o2, (-3*2-1), %o2
	
	
	
LOC(2.15):
	! remainder is negative
	addcc	%o3,%o5,%o3
			! depth 3, accumulated bits -3
	bl	LOC(3.13)
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
			! depth 4, accumulated bits -5
	bl	LOC(4.11)
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
		b	9f
		add	%o2, (-5*2+1), %o2
	
LOC(4.11):
	! remainder is negative
	addcc	%o3,%o5,%o3
		b	9f
		add	%o2, (-5*2-1), %o2
	
	
LOC(3.13):
	! remainder is negative
	addcc	%o3,%o5,%o3
			! depth 4, accumulated bits -7
	bl	LOC(4.9)
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
		b	9f
		add	%o2, (-7*2+1), %o2
	
LOC(4.9):
	! remainder is negative
	addcc	%o3,%o5,%o3
		b	9f
		add	%o2, (-7*2-1), %o2
	
	
	
	
	9:
LOC(end_regular_divide):
	subcc	%o4, 1, %o4
	bge	LOC(divloop)
	tst	%o3
	bl,a	LOC(got_result)
	! non-restoring fixup here (one instruction only!)
	add	%o3, %o1, %o3


LOC(got_result):
	! check to see if answer should be < 0
	tst	%g3
	bl,a	1f
	sub %g0, %o3, %o3
1:
	retl
	mov %o3, %o0

END(.rem)