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Linus Torvalds1da177e2005-04-16 15:20:36 -07001/*
2 * arch/alpha/lib/ev6-clear_user.S
3 * 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com>
4 *
5 * Zero user space, handling exceptions as we go.
6 *
7 * We have to make sure that $0 is always up-to-date and contains the
8 * right "bytes left to zero" value (and that it is updated only _after_
9 * a successful copy). There is also some rather minor exception setup
10 * stuff.
11 *
12 * NOTE! This is not directly C-callable, because the calling semantics
13 * are different:
14 *
15 * Inputs:
16 * length in $0
17 * destination address in $6
18 * exception pointer in $7
19 * return address in $28 (exceptions expect it there)
20 *
21 * Outputs:
22 * bytes left to copy in $0
23 *
24 * Clobbers:
25 * $1,$2,$3,$4,$5,$6
26 *
27 * Much of the information about 21264 scheduling/coding comes from:
28 * Compiler Writer's Guide for the Alpha 21264
29 * abbreviated as 'CWG' in other comments here
30 * ftp.digital.com/pub/Digital/info/semiconductor/literature/dsc-library.html
31 * Scheduling notation:
32 * E - either cluster
33 * U - upper subcluster; U0 - subcluster U0; U1 - subcluster U1
34 * L - lower subcluster; L0 - subcluster L0; L1 - subcluster L1
35 * Try not to change the actual algorithm if possible for consistency.
36 * Determining actual stalls (other than slotting) doesn't appear to be easy to do.
37 * From perusing the source code context where this routine is called, it is
38 * a fair assumption that significant fractions of entire pages are zeroed, so
39 * it's going to be worth the effort to hand-unroll a big loop, and use wh64.
40 * ASSUMPTION:
41 * The believed purpose of only updating $0 after a store is that a signal
42 * may come along during the execution of this chunk of code, and we don't
43 * want to leave a hole (and we also want to avoid repeating lots of work)
44 */
45
46/* Allow an exception for an insn; exit if we get one. */
47#define EX(x,y...) \
48 99: x,##y; \
49 .section __ex_table,"a"; \
50 .long 99b - .; \
51 lda $31, $exception-99b($31); \
52 .previous
53
54 .set noat
55 .set noreorder
56 .align 4
57
58 .globl __do_clear_user
59 .ent __do_clear_user
60 .frame $30, 0, $28
61 .prologue 0
62
63 # Pipeline info : Slotting & Comments
64__do_clear_user:
65 and $6, 7, $4 # .. E .. .. : find dest head misalignment
66 beq $0, $zerolength # U .. .. .. : U L U L
67
68 addq $0, $4, $1 # .. .. .. E : bias counter
69 and $1, 7, $2 # .. .. E .. : number of misaligned bytes in tail
70# Note - we never actually use $2, so this is a moot computation
71# and we can rewrite this later...
72 srl $1, 3, $1 # .. E .. .. : number of quadwords to clear
73 beq $4, $headalign # U .. .. .. : U L U L
74
75/*
76 * Head is not aligned. Write (8 - $4) bytes to head of destination
77 * This means $6 is known to be misaligned
78 */
79 EX( ldq_u $5, 0($6) ) # .. .. .. L : load dst word to mask back in
80 beq $1, $onebyte # .. .. U .. : sub-word store?
81 mskql $5, $6, $5 # .. U .. .. : take care of misaligned head
82 addq $6, 8, $6 # E .. .. .. : L U U L
83
84 EX( stq_u $5, -8($6) ) # .. .. .. L :
85 subq $1, 1, $1 # .. .. E .. :
86 addq $0, $4, $0 # .. E .. .. : bytes left -= 8 - misalignment
87 subq $0, 8, $0 # E .. .. .. : U L U L
88
89 .align 4
90/*
91 * (The .align directive ought to be a moot point)
92 * values upon initial entry to the loop
93 * $1 is number of quadwords to clear (zero is a valid value)
94 * $2 is number of trailing bytes (0..7) ($2 never used...)
95 * $6 is known to be aligned 0mod8
96 */
97$headalign:
98 subq $1, 16, $4 # .. .. .. E : If < 16, we can not use the huge loop
99 and $6, 0x3f, $2 # .. .. E .. : Forward work for huge loop
100 subq $2, 0x40, $3 # .. E .. .. : bias counter (huge loop)
101 blt $4, $trailquad # U .. .. .. : U L U L
102
103/*
104 * We know that we're going to do at least 16 quads, which means we are
105 * going to be able to use the large block clear loop at least once.
106 * Figure out how many quads we need to clear before we are 0mod64 aligned
107 * so we can use the wh64 instruction.
108 */
109
110 nop # .. .. .. E
111 nop # .. .. E ..
112 nop # .. E .. ..
113 beq $3, $bigalign # U .. .. .. : U L U L : Aligned 0mod64
114
115$alignmod64:
116 EX( stq_u $31, 0($6) ) # .. .. .. L
117 addq $3, 8, $3 # .. .. E ..
118 subq $0, 8, $0 # .. E .. ..
119 nop # E .. .. .. : U L U L
120
121 nop # .. .. .. E
122 subq $1, 1, $1 # .. .. E ..
123 addq $6, 8, $6 # .. E .. ..
124 blt $3, $alignmod64 # U .. .. .. : U L U L
125
126$bigalign:
127/*
128 * $0 is the number of bytes left
129 * $1 is the number of quads left
130 * $6 is aligned 0mod64
131 * we know that we'll be taking a minimum of one trip through
132 * CWG Section 3.7.6: do not expect a sustained store rate of > 1/cycle
133 * We are _not_ going to update $0 after every single store. That
134 * would be silly, because there will be cross-cluster dependencies
135 * no matter how the code is scheduled. By doing it in slightly
136 * staggered fashion, we can still do this loop in 5 fetches
137 * The worse case will be doing two extra quads in some future execution,
138 * in the event of an interrupted clear.
139 * Assumes the wh64 needs to be for 2 trips through the loop in the future
140 * The wh64 is issued on for the starting destination address for trip +2
141 * through the loop, and if there are less than two trips left, the target
142 * address will be for the current trip.
143 */
144 nop # E :
145 nop # E :
146 nop # E :
147 bis $6,$6,$3 # E : U L U L : Initial wh64 address is dest
148 /* This might actually help for the current trip... */
149
150$do_wh64:
151 wh64 ($3) # .. .. .. L1 : memory subsystem hint
152 subq $1, 16, $4 # .. .. E .. : Forward calculation - repeat the loop?
153 EX( stq_u $31, 0($6) ) # .. L .. ..
154 subq $0, 8, $0 # E .. .. .. : U L U L
155
156 addq $6, 128, $3 # E : Target address of wh64
157 EX( stq_u $31, 8($6) ) # L :
158 EX( stq_u $31, 16($6) ) # L :
159 subq $0, 16, $0 # E : U L L U
160
161 nop # E :
162 EX( stq_u $31, 24($6) ) # L :
163 EX( stq_u $31, 32($6) ) # L :
164 subq $0, 168, $5 # E : U L L U : two trips through the loop left?
165 /* 168 = 192 - 24, since we've already completed some stores */
166
167 subq $0, 16, $0 # E :
168 EX( stq_u $31, 40($6) ) # L :
169 EX( stq_u $31, 48($6) ) # L :
170 cmovlt $5, $6, $3 # E : U L L U : Latency 2, extra mapping cycle
171
172 subq $1, 8, $1 # E :
173 subq $0, 16, $0 # E :
174 EX( stq_u $31, 56($6) ) # L :
175 nop # E : U L U L
176
177 nop # E :
178 subq $0, 8, $0 # E :
179 addq $6, 64, $6 # E :
180 bge $4, $do_wh64 # U : U L U L
181
182$trailquad:
183 # zero to 16 quadwords left to store, plus any trailing bytes
184 # $1 is the number of quadwords left to go.
185 #
186 nop # .. .. .. E
187 nop # .. .. E ..
188 nop # .. E .. ..
189 beq $1, $trailbytes # U .. .. .. : U L U L : Only 0..7 bytes to go
190
191$onequad:
192 EX( stq_u $31, 0($6) ) # .. .. .. L
193 subq $1, 1, $1 # .. .. E ..
194 subq $0, 8, $0 # .. E .. ..
195 nop # E .. .. .. : U L U L
196
197 nop # .. .. .. E
198 nop # .. .. E ..
199 addq $6, 8, $6 # .. E .. ..
200 bgt $1, $onequad # U .. .. .. : U L U L
201
202 # We have an unknown number of bytes left to go.
203$trailbytes:
204 nop # .. .. .. E
205 nop # .. .. E ..
206 nop # .. E .. ..
207 beq $0, $zerolength # U .. .. .. : U L U L
208
209 # $0 contains the number of bytes left to copy (0..31)
210 # so we will use $0 as the loop counter
211 # We know for a fact that $0 > 0 zero due to previous context
212$onebyte:
213 EX( stb $31, 0($6) ) # .. .. .. L
214 subq $0, 1, $0 # .. .. E .. :
215 addq $6, 1, $6 # .. E .. .. :
216 bgt $0, $onebyte # U .. .. .. : U L U L
217
218$zerolength:
219$exception: # Destination for exception recovery(?)
220 nop # .. .. .. E :
221 nop # .. .. E .. :
222 nop # .. E .. .. :
223 ret $31, ($28), 1 # L0 .. .. .. : L U L U
224 .end __do_clear_user
225