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b2441318 | 1 | /* SPDX-License-Identifier: GPL-2.0 */ |
1da177e4 LT |
2 | /* |
3 | * arch/alpha/lib/ev6-clear_user.S | |
4 | * 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com> | |
5 | * | |
6 | * Zero user space, handling exceptions as we go. | |
7 | * | |
8 | * We have to make sure that $0 is always up-to-date and contains the | |
9 | * right "bytes left to zero" value (and that it is updated only _after_ | |
10 | * a successful copy). There is also some rather minor exception setup | |
11 | * stuff. | |
12 | * | |
1da177e4 LT |
13 | * Much of the information about 21264 scheduling/coding comes from: |
14 | * Compiler Writer's Guide for the Alpha 21264 | |
15 | * abbreviated as 'CWG' in other comments here | |
16 | * ftp.digital.com/pub/Digital/info/semiconductor/literature/dsc-library.html | |
17 | * Scheduling notation: | |
18 | * E - either cluster | |
19 | * U - upper subcluster; U0 - subcluster U0; U1 - subcluster U1 | |
20 | * L - lower subcluster; L0 - subcluster L0; L1 - subcluster L1 | |
21 | * Try not to change the actual algorithm if possible for consistency. | |
22 | * Determining actual stalls (other than slotting) doesn't appear to be easy to do. | |
23 | * From perusing the source code context where this routine is called, it is | |
24 | * a fair assumption that significant fractions of entire pages are zeroed, so | |
25 | * it's going to be worth the effort to hand-unroll a big loop, and use wh64. | |
26 | * ASSUMPTION: | |
27 | * The believed purpose of only updating $0 after a store is that a signal | |
28 | * may come along during the execution of this chunk of code, and we don't | |
29 | * want to leave a hole (and we also want to avoid repeating lots of work) | |
30 | */ | |
31 | ||
00fc0e0d | 32 | #include <asm/export.h> |
1da177e4 LT |
33 | /* Allow an exception for an insn; exit if we get one. */ |
34 | #define EX(x,y...) \ | |
35 | 99: x,##y; \ | |
36 | .section __ex_table,"a"; \ | |
37 | .long 99b - .; \ | |
38 | lda $31, $exception-99b($31); \ | |
39 | .previous | |
40 | ||
41 | .set noat | |
42 | .set noreorder | |
43 | .align 4 | |
44 | ||
85250231 AV |
45 | .globl __clear_user |
46 | .ent __clear_user | |
47 | .frame $30, 0, $26 | |
1da177e4 LT |
48 | .prologue 0 |
49 | ||
50 | # Pipeline info : Slotting & Comments | |
85250231 AV |
51 | __clear_user: |
52 | and $17, $17, $0 | |
53 | and $16, 7, $4 # .. E .. .. : find dest head misalignment | |
1da177e4 LT |
54 | beq $0, $zerolength # U .. .. .. : U L U L |
55 | ||
56 | addq $0, $4, $1 # .. .. .. E : bias counter | |
57 | and $1, 7, $2 # .. .. E .. : number of misaligned bytes in tail | |
58 | # Note - we never actually use $2, so this is a moot computation | |
59 | # and we can rewrite this later... | |
60 | srl $1, 3, $1 # .. E .. .. : number of quadwords to clear | |
61 | beq $4, $headalign # U .. .. .. : U L U L | |
62 | ||
63 | /* | |
64 | * Head is not aligned. Write (8 - $4) bytes to head of destination | |
85250231 | 65 | * This means $16 is known to be misaligned |
1da177e4 | 66 | */ |
85250231 | 67 | EX( ldq_u $5, 0($16) ) # .. .. .. L : load dst word to mask back in |
1da177e4 | 68 | beq $1, $onebyte # .. .. U .. : sub-word store? |
85250231 AV |
69 | mskql $5, $16, $5 # .. U .. .. : take care of misaligned head |
70 | addq $16, 8, $16 # E .. .. .. : L U U L | |
1da177e4 | 71 | |
85250231 | 72 | EX( stq_u $5, -8($16) ) # .. .. .. L : |
1da177e4 LT |
73 | subq $1, 1, $1 # .. .. E .. : |
74 | addq $0, $4, $0 # .. E .. .. : bytes left -= 8 - misalignment | |
75 | subq $0, 8, $0 # E .. .. .. : U L U L | |
76 | ||
77 | .align 4 | |
78 | /* | |
79 | * (The .align directive ought to be a moot point) | |
80 | * values upon initial entry to the loop | |
81 | * $1 is number of quadwords to clear (zero is a valid value) | |
82 | * $2 is number of trailing bytes (0..7) ($2 never used...) | |
85250231 | 83 | * $16 is known to be aligned 0mod8 |
1da177e4 LT |
84 | */ |
85 | $headalign: | |
86 | subq $1, 16, $4 # .. .. .. E : If < 16, we can not use the huge loop | |
85250231 | 87 | and $16, 0x3f, $2 # .. .. E .. : Forward work for huge loop |
1da177e4 LT |
88 | subq $2, 0x40, $3 # .. E .. .. : bias counter (huge loop) |
89 | blt $4, $trailquad # U .. .. .. : U L U L | |
90 | ||
91 | /* | |
92 | * We know that we're going to do at least 16 quads, which means we are | |
93 | * going to be able to use the large block clear loop at least once. | |
94 | * Figure out how many quads we need to clear before we are 0mod64 aligned | |
95 | * so we can use the wh64 instruction. | |
96 | */ | |
97 | ||
98 | nop # .. .. .. E | |
99 | nop # .. .. E .. | |
100 | nop # .. E .. .. | |
101 | beq $3, $bigalign # U .. .. .. : U L U L : Aligned 0mod64 | |
102 | ||
103 | $alignmod64: | |
85250231 | 104 | EX( stq_u $31, 0($16) ) # .. .. .. L |
1da177e4 LT |
105 | addq $3, 8, $3 # .. .. E .. |
106 | subq $0, 8, $0 # .. E .. .. | |
107 | nop # E .. .. .. : U L U L | |
108 | ||
109 | nop # .. .. .. E | |
110 | subq $1, 1, $1 # .. .. E .. | |
85250231 | 111 | addq $16, 8, $16 # .. E .. .. |
1da177e4 LT |
112 | blt $3, $alignmod64 # U .. .. .. : U L U L |
113 | ||
114 | $bigalign: | |
115 | /* | |
116 | * $0 is the number of bytes left | |
117 | * $1 is the number of quads left | |
85250231 | 118 | * $16 is aligned 0mod64 |
1da177e4 LT |
119 | * we know that we'll be taking a minimum of one trip through |
120 | * CWG Section 3.7.6: do not expect a sustained store rate of > 1/cycle | |
121 | * We are _not_ going to update $0 after every single store. That | |
122 | * would be silly, because there will be cross-cluster dependencies | |
123 | * no matter how the code is scheduled. By doing it in slightly | |
124 | * staggered fashion, we can still do this loop in 5 fetches | |
125 | * The worse case will be doing two extra quads in some future execution, | |
126 | * in the event of an interrupted clear. | |
127 | * Assumes the wh64 needs to be for 2 trips through the loop in the future | |
128 | * The wh64 is issued on for the starting destination address for trip +2 | |
129 | * through the loop, and if there are less than two trips left, the target | |
130 | * address will be for the current trip. | |
131 | */ | |
132 | nop # E : | |
133 | nop # E : | |
134 | nop # E : | |
85250231 | 135 | bis $16,$16,$3 # E : U L U L : Initial wh64 address is dest |
1da177e4 LT |
136 | /* This might actually help for the current trip... */ |
137 | ||
138 | $do_wh64: | |
139 | wh64 ($3) # .. .. .. L1 : memory subsystem hint | |
140 | subq $1, 16, $4 # .. .. E .. : Forward calculation - repeat the loop? | |
85250231 | 141 | EX( stq_u $31, 0($16) ) # .. L .. .. |
1da177e4 LT |
142 | subq $0, 8, $0 # E .. .. .. : U L U L |
143 | ||
85250231 AV |
144 | addq $16, 128, $3 # E : Target address of wh64 |
145 | EX( stq_u $31, 8($16) ) # L : | |
146 | EX( stq_u $31, 16($16) ) # L : | |
1da177e4 LT |
147 | subq $0, 16, $0 # E : U L L U |
148 | ||
149 | nop # E : | |
85250231 AV |
150 | EX( stq_u $31, 24($16) ) # L : |
151 | EX( stq_u $31, 32($16) ) # L : | |
1da177e4 LT |
152 | subq $0, 168, $5 # E : U L L U : two trips through the loop left? |
153 | /* 168 = 192 - 24, since we've already completed some stores */ | |
154 | ||
155 | subq $0, 16, $0 # E : | |
85250231 AV |
156 | EX( stq_u $31, 40($16) ) # L : |
157 | EX( stq_u $31, 48($16) ) # L : | |
158 | cmovlt $5, $16, $3 # E : U L L U : Latency 2, extra mapping cycle | |
1da177e4 LT |
159 | |
160 | subq $1, 8, $1 # E : | |
161 | subq $0, 16, $0 # E : | |
85250231 | 162 | EX( stq_u $31, 56($16) ) # L : |
1da177e4 LT |
163 | nop # E : U L U L |
164 | ||
165 | nop # E : | |
166 | subq $0, 8, $0 # E : | |
85250231 | 167 | addq $16, 64, $16 # E : |
1da177e4 LT |
168 | bge $4, $do_wh64 # U : U L U L |
169 | ||
170 | $trailquad: | |
171 | # zero to 16 quadwords left to store, plus any trailing bytes | |
172 | # $1 is the number of quadwords left to go. | |
173 | # | |
174 | nop # .. .. .. E | |
175 | nop # .. .. E .. | |
176 | nop # .. E .. .. | |
177 | beq $1, $trailbytes # U .. .. .. : U L U L : Only 0..7 bytes to go | |
178 | ||
179 | $onequad: | |
85250231 | 180 | EX( stq_u $31, 0($16) ) # .. .. .. L |
1da177e4 LT |
181 | subq $1, 1, $1 # .. .. E .. |
182 | subq $0, 8, $0 # .. E .. .. | |
183 | nop # E .. .. .. : U L U L | |
184 | ||
185 | nop # .. .. .. E | |
186 | nop # .. .. E .. | |
85250231 | 187 | addq $16, 8, $16 # .. E .. .. |
1da177e4 LT |
188 | bgt $1, $onequad # U .. .. .. : U L U L |
189 | ||
190 | # We have an unknown number of bytes left to go. | |
191 | $trailbytes: | |
192 | nop # .. .. .. E | |
193 | nop # .. .. E .. | |
194 | nop # .. E .. .. | |
195 | beq $0, $zerolength # U .. .. .. : U L U L | |
196 | ||
197 | # $0 contains the number of bytes left to copy (0..31) | |
198 | # so we will use $0 as the loop counter | |
199 | # We know for a fact that $0 > 0 zero due to previous context | |
200 | $onebyte: | |
85250231 | 201 | EX( stb $31, 0($16) ) # .. .. .. L |
1da177e4 | 202 | subq $0, 1, $0 # .. .. E .. : |
85250231 | 203 | addq $16, 1, $16 # .. E .. .. : |
1da177e4 LT |
204 | bgt $0, $onebyte # U .. .. .. : U L U L |
205 | ||
206 | $zerolength: | |
207 | $exception: # Destination for exception recovery(?) | |
208 | nop # .. .. .. E : | |
209 | nop # .. .. E .. : | |
210 | nop # .. E .. .. : | |
85250231 AV |
211 | ret $31, ($26), 1 # L0 .. .. .. : L U L U |
212 | .end __clear_user | |
213 | EXPORT_SYMBOL(__clear_user) |