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1394f032 | 1 | /* |
96f1050d | 2 | * Copyright 2004-2009 Analog Devices Inc. |
1394f032 | 3 | * |
de450838 | 4 | * Licensed under the Clear BSD license or the GPL-2 (or later) |
1394f032 BW |
5 | */ |
6 | ||
7 | #include <linux/linkage.h> | |
8 | ||
9 | #define CARRY AC0 | |
10 | ||
11 | #ifdef CONFIG_ARITHMETIC_OPS_L1 | |
12 | .section .l1.text | |
13 | #else | |
14 | .text | |
15 | #endif | |
16 | ||
17 | ||
18 | ENTRY(___udivsi3) | |
19 | ||
20 | CC = R0 < R1 (IU); /* If X < Y, always return 0 */ | |
21 | IF CC JUMP .Lreturn_ident; | |
22 | ||
23 | R2 = R1 << 16; | |
24 | CC = R2 <= R0 (IU); | |
25 | IF CC JUMP .Lidents; | |
26 | ||
27 | R2 = R0 >> 31; /* if X is a 31-bit number */ | |
28 | R3 = R1 >> 15; /* and Y is a 15-bit number */ | |
29 | R2 = R2 | R3; /* then it's okay to use the DIVQ builtins (fallthrough to fast)*/ | |
30 | CC = R2; | |
31 | IF CC JUMP .Ly_16bit; | |
32 | ||
33 | /* METHOD 1: FAST DIVQ | |
34 | We know we have a 31-bit dividend, and 15-bit divisor so we can use the | |
35 | simple divq approach (first setting AQ to 0 - implying unsigned division, | |
36 | then 16 DIVQ's). | |
37 | */ | |
38 | ||
39 | AQ = CC; /* Clear AQ (CC==0) */ | |
40 | ||
41 | /* ISR States: When dividing two integers (32.0/16.0) using divide primitives, | |
42 | we need to shift the dividend one bit to the left. | |
43 | We have already checked that we have a 31-bit number so we are safe to do | |
44 | that. | |
45 | */ | |
46 | R0 <<= 1; | |
47 | DIVQ(R0, R1); // 1 | |
48 | DIVQ(R0, R1); // 2 | |
49 | DIVQ(R0, R1); // 3 | |
50 | DIVQ(R0, R1); // 4 | |
51 | DIVQ(R0, R1); // 5 | |
52 | DIVQ(R0, R1); // 6 | |
53 | DIVQ(R0, R1); // 7 | |
54 | DIVQ(R0, R1); // 8 | |
55 | DIVQ(R0, R1); // 9 | |
56 | DIVQ(R0, R1); // 10 | |
57 | DIVQ(R0, R1); // 11 | |
58 | DIVQ(R0, R1); // 12 | |
59 | DIVQ(R0, R1); // 13 | |
60 | DIVQ(R0, R1); // 14 | |
61 | DIVQ(R0, R1); // 15 | |
62 | DIVQ(R0, R1); // 16 | |
63 | R0 = R0.L (Z); | |
64 | RTS; | |
65 | ||
66 | .Ly_16bit: | |
67 | /* We know that the upper 17 bits of Y might have bits set, | |
68 | ** or that the sign bit of X might have a bit. If Y is a | |
69 | ** 16-bit number, but not bigger, then we can use the builtins | |
70 | ** with a post-divide correction. | |
71 | ** R3 currently holds Y>>15, which means R3's LSB is the | |
72 | ** bit we're interested in. | |
73 | */ | |
74 | ||
75 | /* According to the ISR, to use the Divide primitives for | |
76 | ** unsigned integer divide, the useable range is 31 bits | |
77 | */ | |
78 | CC = ! BITTST(R0, 31); | |
79 | ||
80 | /* IF condition is true we can scale our inputs and use the divide primitives, | |
81 | ** with some post-adjustment | |
82 | */ | |
83 | R3 += -1; /* if so, Y is 0x00008nnn */ | |
84 | CC &= AZ; | |
85 | ||
86 | /* If condition is true we can scale our inputs and use the divide primitives, | |
87 | ** with some post-adjustment | |
88 | */ | |
89 | R3 = R1 >> 1; /* Pre-scaled divisor for primitive case */ | |
90 | R2 = R0 >> 16; | |
91 | ||
92 | R2 = R3 - R2; /* shifted divisor < upper 16 bits of dividend */ | |
93 | CC &= CARRY; | |
94 | IF CC JUMP .Lshift_and_correct; | |
95 | ||
96 | /* Fall through to the identities */ | |
97 | ||
98 | /* METHOD 2: identities and manual calculation | |
99 | We are not able to use the divide primites, but may still catch some special | |
100 | cases. | |
101 | */ | |
102 | .Lidents: | |
103 | /* Test for common identities. Value to be returned is placed in R2. */ | |
104 | CC = R0 == 0; /* 0/Y => 0 */ | |
105 | IF CC JUMP .Lreturn_r0; | |
106 | CC = R0 == R1; /* X==Y => 1 */ | |
107 | IF CC JUMP .Lreturn_ident; | |
108 | CC = R1 == 1; /* X/1 => X */ | |
109 | IF CC JUMP .Lreturn_ident; | |
110 | ||
111 | R2.L = ONES R1; | |
112 | R2 = R2.L (Z); | |
113 | CC = R2 == 1; | |
114 | IF CC JUMP .Lpower_of_two; | |
115 | ||
116 | [--SP] = (R7:5); /* Push registers R5-R7 */ | |
117 | ||
118 | /* Idents don't match. Go for the full operation. */ | |
119 | ||
120 | ||
121 | R6 = 2; /* assume we'll shift two */ | |
122 | R3 = 1; | |
123 | ||
124 | P2 = R1; | |
125 | /* If either R0 or R1 have sign set, */ | |
126 | /* divide them by two, and note it's */ | |
127 | /* been done. */ | |
128 | CC = R1 < 0; | |
129 | R2 = R1 >> 1; | |
130 | IF CC R1 = R2; /* Possibly-shifted R1 */ | |
131 | IF !CC R6 = R3; /* R1 doesn't, so at most 1 shifted */ | |
132 | ||
133 | P0 = 0; | |
134 | R3 = -R1; | |
135 | [--SP] = R3; | |
136 | R2 = R0 >> 1; | |
137 | R2 = R0 >> 1; | |
138 | CC = R0 < 0; | |
139 | IF CC P0 = R6; /* Number of values divided */ | |
140 | IF !CC R2 = R0; /* Shifted R0 */ | |
141 | ||
142 | /* P0 is 0, 1 (NR/=2) or 2 (NR/=2, DR/=2) */ | |
143 | ||
144 | /* r2 holds Copy dividend */ | |
145 | R3 = 0; /* Clear partial remainder */ | |
146 | R7 = 0; /* Initialise quotient bit */ | |
147 | ||
148 | P1 = 32; /* Set loop counter */ | |
149 | LSETUP(.Lulst, .Lulend) LC0 = P1; /* Set loop counter */ | |
150 | .Lulst: R6 = R2 >> 31; /* R6 = sign bit of R2, for carry */ | |
151 | R2 = R2 << 1; /* Shift 64 bit dividend up by 1 bit */ | |
152 | R3 = R3 << 1 || R5 = [SP]; | |
153 | R3 = R3 | R6; /* Include any carry */ | |
154 | CC = R7 < 0; /* Check quotient(AQ) */ | |
155 | /* If AQ==0, we'll sub divisor */ | |
156 | IF CC R5 = R1; /* and if AQ==1, we'll add it. */ | |
bc2d52fe | 157 | R3 = R3 + R5; /* Add/sub divisor to partial remainder */ |
1394f032 BW |
158 | R7 = R3 ^ R1; /* Generate next quotient bit */ |
159 | ||
160 | R5 = R7 >> 31; /* Get AQ */ | |
161 | BITTGL(R5, 0); /* Invert it, to get what we'll shift */ | |
162 | .Lulend: R2 = R2 + R5; /* and "shift" it in. */ | |
163 | ||
164 | CC = P0 == 0; /* Check how many inputs we shifted */ | |
165 | IF CC JUMP .Lno_mult; /* if none... */ | |
166 | R6 = R2 << 1; | |
167 | CC = P0 == 1; | |
168 | IF CC R2 = R6; /* if 1, Q = Q*2 */ | |
169 | IF !CC R1 = P2; /* if 2, restore stored divisor */ | |
170 | ||
171 | R3 = R2; /* Copy of R2 */ | |
172 | R3 *= R1; /* Q * divisor */ | |
173 | R5 = R0 - R3; /* Z = (dividend - Q * divisor) */ | |
174 | CC = R1 <= R5 (IU); /* Check if divisor <= Z? */ | |
175 | R6 = CC; /* if yes, R6 = 1 */ | |
176 | R2 = R2 + R6; /* if yes, add one to quotient(Q) */ | |
177 | .Lno_mult: | |
178 | SP += 4; | |
179 | (R7:5) = [SP++]; /* Pop registers R5-R7 */ | |
180 | R0 = R2; /* Store quotient */ | |
181 | RTS; | |
182 | ||
183 | .Lreturn_ident: | |
184 | CC = R0 < R1 (IU); /* If X < Y, always return 0 */ | |
185 | R2 = 0; | |
186 | IF CC JUMP .Ltrue_return_ident; | |
187 | R2 = -1 (X); /* X/0 => 0xFFFFFFFF */ | |
188 | CC = R1 == 0; | |
189 | IF CC JUMP .Ltrue_return_ident; | |
190 | R2 = -R2; /* R2 now 1 */ | |
191 | CC = R0 == R1; /* X==Y => 1 */ | |
192 | IF CC JUMP .Ltrue_return_ident; | |
193 | R2 = R0; /* X/1 => X */ | |
194 | /*FALLTHRU*/ | |
195 | ||
196 | .Ltrue_return_ident: | |
197 | R0 = R2; | |
198 | .Lreturn_r0: | |
199 | RTS; | |
200 | ||
201 | .Lpower_of_two: | |
202 | /* Y has a single bit set, which means it's a power of two. | |
203 | ** That means we can perform the division just by shifting | |
204 | ** X to the right the appropriate number of bits | |
205 | */ | |
206 | ||
207 | /* signbits returns the number of sign bits, minus one. | |
208 | ** 1=>30, 2=>29, ..., 0x40000000=>0. Which means we need | |
209 | ** to shift right n-signbits spaces. It also means 0x80000000 | |
210 | ** is a special case, because that *also* gives a signbits of 0 | |
211 | */ | |
212 | ||
213 | R2 = R0 >> 31; | |
214 | CC = R1 < 0; | |
215 | IF CC JUMP .Ltrue_return_ident; | |
216 | ||
217 | R1.l = SIGNBITS R1; | |
218 | R1 = R1.L (Z); | |
219 | R1 += -30; | |
220 | R0 = LSHIFT R0 by R1.L; | |
221 | RTS; | |
222 | ||
223 | /* METHOD 3: PRESCALE AND USE THE DIVIDE PRIMITIVES WITH SOME POST-CORRECTION | |
224 | Two scaling operations are required to use the divide primitives with a | |
225 | divisor > 0x7FFFF. | |
226 | Firstly (as in method 1) we need to shift the dividend 1 to the left for | |
227 | integer division. | |
228 | Secondly we need to shift both the divisor and dividend 1 to the right so | |
229 | both are in range for the primitives. | |
230 | The left/right shift of the dividend does nothing so we can skip it. | |
231 | */ | |
232 | .Lshift_and_correct: | |
233 | R2 = R0; | |
234 | // R3 is already R1 >> 1 | |
235 | CC=!CC; | |
236 | AQ = CC; /* Clear AQ, got here with CC = 0 */ | |
237 | DIVQ(R2, R3); // 1 | |
238 | DIVQ(R2, R3); // 2 | |
239 | DIVQ(R2, R3); // 3 | |
240 | DIVQ(R2, R3); // 4 | |
241 | DIVQ(R2, R3); // 5 | |
242 | DIVQ(R2, R3); // 6 | |
243 | DIVQ(R2, R3); // 7 | |
244 | DIVQ(R2, R3); // 8 | |
245 | DIVQ(R2, R3); // 9 | |
246 | DIVQ(R2, R3); // 10 | |
247 | DIVQ(R2, R3); // 11 | |
248 | DIVQ(R2, R3); // 12 | |
249 | DIVQ(R2, R3); // 13 | |
250 | DIVQ(R2, R3); // 14 | |
251 | DIVQ(R2, R3); // 15 | |
252 | DIVQ(R2, R3); // 16 | |
253 | ||
254 | /* According to the Instruction Set Reference: | |
255 | To divide by a divisor > 0x7FFF, | |
256 | 1. prescale and perform divide to obtain quotient (Q) (done above), | |
257 | 2. multiply quotient by unscaled divisor (result M) | |
258 | 3. subtract the product from the divident to get an error (E = X - M) | |
259 | 4. if E < divisor (Y) subtract 1, if E > divisor (Y) add 1, else return quotient (Q) | |
260 | */ | |
261 | R3 = R2.L (Z); /* Q = X' / Y' */ | |
262 | R2 = R3; /* Preserve Q */ | |
263 | R2 *= R1; /* M = Q * Y */ | |
264 | R2 = R0 - R2; /* E = X - M */ | |
265 | R0 = R3; /* Copy Q into result reg */ | |
266 | ||
267 | /* Correction: If result of the multiply is negative, we overflowed | |
268 | and need to correct the result by subtracting 1 from the result.*/ | |
269 | R3 = 0xFFFF (Z); | |
270 | R2 = R2 >> 16; /* E >> 16 */ | |
271 | CC = R2 == R3; | |
272 | R3 = 1 ; | |
273 | R1 = R0 - R3; | |
274 | IF CC R0 = R1; | |
275 | RTS; | |
51be24c3 MF |
276 | |
277 | ENDPROC(___udivsi3) |