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
|
/*
* 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:
* NAME name of function to generate
* OP OP=div => %o0 / %o1; OP=rem => %o0 % %o1
* S S=true => signed; S=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.
*/
define(N, `4')dnl
define(WORDSIZE, `32')dnl
define(TOPBITS, eval(WORDSIZE - N*((WORDSIZE-1)/N)))dnl
dnl
define(dividend, `%o0')dnl
define(divisor, `%o1')dnl
define(Q, `%o2')dnl
define(R, `%o3')dnl
define(ITER, `%o4')dnl
define(V, `%o5')dnl
dnl
dnl m4 reminder: ifelse(a,b,c,d) => if a is b, then c, else d
define(T, `%g1')dnl
define(SC, `%g7')dnl
ifelse(S, `true', `define(SIGN, `%g6')')dnl
dnl
dnl This is the recursive definition for developing quotient digits.
dnl
dnl Parameters:
dnl $1 the current depth, 1 <= $1 <= N
dnl $2 the current accumulation of quotient bits
dnl N max depth
dnl
dnl We add a new bit to $2 and either recurse or insert the bits in
dnl the quotient. R, Q, and V are inputs and outputs as defined above;
dnl the condition codes are expected to reflect the input R, and are
dnl modified to reflect the output R.
dnl
define(DEVELOP_QUOTIENT_BITS,
` ! depth $1, accumulated bits $2
bl L.$1.eval(2**N+$2)
srl V,1,V
! remainder is positive
subcc R,V,R
ifelse($1, N,
` b 9f
add Q, ($2*2+1), Q
', ` DEVELOP_QUOTIENT_BITS(incr($1), `eval(2*$2+1)')')
L.$1.eval(2**N+$2):
! remainder is negative
addcc R,V,R
ifelse($1, N,
` b 9f
add Q, ($2*2-1), Q
', ` DEVELOP_QUOTIENT_BITS(incr($1), `eval(2*$2-1)')')
ifelse($1, 1, `9:')')dnl
#include "DEFS.h"
#ifdef __svr4__
#include <sys/trap.h>
#else
#include <machine/trap.h>
#endif
FUNC(NAME)
ifelse(S, `true',
` ! compute sign of result; if neither is negative, no problem
orcc divisor, dividend, %g0 ! either negative?
bge 2f ! no, go do the divide
ifelse(OP, `div',
` xor divisor, dividend, SIGN ! compute sign in any case',
` mov dividend, SIGN ! sign of remainder matches dividend')
tst divisor
bge 1f
tst dividend
! divisor is definitely negative; dividend might also be negative
bge 2f ! if dividend not negative...
sub %g0, divisor, divisor ! in any case, make divisor nonneg
1: ! dividend is negative, divisor is nonnegative
sub %g0, dividend, dividend ! make dividend nonnegative
2:
')
! Ready to divide. Compute size of quotient; scale comparand.
orcc divisor, %g0, V
bne 1f
mov dividend, R
! 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 R, V ! if divisor exceeds dividend, done
blu Lgot_result ! (and algorithm fails otherwise)
clr Q
sethi %hi(1 << (WORDSIZE - TOPBITS - 1)), T
cmp R, T
blu Lnot_really_big
clr ITER
! `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 V, T
bgeu 3f
mov 1, SC
sll V, N, V
b 1b
add ITER, 1, ITER
! Now compute SC.
2: addcc V, V, V
bcc Lnot_too_big
add SC, 1, SC
! We get here if the divisor overflowed while shifting.
! This means that R has the high-order bit set.
! Restore V and subtract from R.
sll T, TOPBITS, T ! high order bit
srl V, 1, V ! rest of V
add V, T, V
b Ldo_single_div
sub SC, 1, SC
Lnot_too_big:
3: cmp V, R
blu 2b
nop
be Ldo_single_div
nop
/* NB: these are commented out in the V8-Sparc manual as well */
/* (I do not understand this) */
! V > R: went too far: back up 1 step
! srl V, 1, V
! dec SC
! do single-bit divide steps
!
! We have to be careful here. We know that R >= V, so we can do the
! first divide step without thinking. BUT, the others are conditional,
! and are only done if R >= 0. Because both R and V 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...
Ldo_single_div:
subcc SC, 1, SC
bl Lend_regular_divide
nop
sub R, V, R
mov 1, Q
b Lend_single_divloop
nop
Lsingle_divloop:
sll Q, 1, Q
bl 1f
srl V, 1, V
! R >= 0
sub R, V, R
b 2f
add Q, 1, Q
1: ! R < 0
add R, V, R
sub Q, 1, Q
2:
Lend_single_divloop:
subcc SC, 1, SC
bge Lsingle_divloop
tst R
b,a Lend_regular_divide
Lnot_really_big:
1:
sll V, N, V
cmp V, R
bleu 1b
addcc ITER, 1, ITER
be Lgot_result
sub ITER, 1, ITER
tst R ! set up for initial iteration
Ldivloop:
sll Q, N, Q
DEVELOP_QUOTIENT_BITS(1, 0)
Lend_regular_divide:
subcc ITER, 1, ITER
bge Ldivloop
tst R
bl,a Lgot_result
! non-restoring fixup here (one instruction only!)
ifelse(OP, `div',
` sub Q, 1, Q
', ` add R, divisor, R
')
Lgot_result:
ifelse(S, `true',
` ! check to see if answer should be < 0
tst SIGN
bl,a 1f
ifelse(OP, `div', `sub %g0, Q, Q', `sub %g0, R, R')
1:')
retl
ifelse(OP, `div', `mov Q, %o0', `mov R, %o0')
|
