1 /* SPDX-License-Identifier: GPL-2.0 */
2 /*
3 * arch/alpha/lib/ev6-clear_user.S
4 * 21264 version contributed by Rick Gorton <ri
5 *
6 * Zero user space, handling exceptions as we
7 *
8 * We have to make sure that $0 is always up-t
9 * right "bytes left to zero" value (and that
10 * a successful copy). There is also some rat
11 * stuff.
12 *
13 * Much of the information about 21264 schedul
14 * Compiler Writer's Guide for the Alpha
15 * abbreviated as 'CWG' in other comments
16 * ftp.digital.com/pub/Digital/info/semic
17 * Scheduling notation:
18 * E - either cluster
19 * U - upper subcluster; U0 - subcl
20 * L - lower subcluster; L0 - subcl
21 * Try not to change the actual algorithm if p
22 * Determining actual stalls (other than slott
23 * From perusing the source code context where
24 * a fair assumption that significant fraction
25 * it's going to be worth the effort to hand-u
26 * ASSUMPTION:
27 * The believed purpose of only updating
28 * may come along during the execution of
29 * want to leave a hole (and we also want
30 */
31
32 #include <linux/export.h>
33 /* Allow an exception for an insn; exit if we
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
45 .globl __clear_user
46 .ent __clear_user
47 .frame $30, 0, $26
48 .prologue 0
49
50 # Pipeline inf
51 __clear_user:
52 and $17, $17, $0
53 and $16, 7, $4 # .. E .. ..
54 beq $0, $zerolength # U .. .. ..
55
56 addq $0, $4, $1 # .. .. .. E
57 and $1, 7, $2 # .. .. E ..
58 # Note - we never actually use $2, so this is
59 # and we can rewrite this later...
60 srl $1, 3, $1 # .. E .. ..
61 beq $4, $headalign # U .. .. ..
62
63 /*
64 * Head is not aligned. Write (8 - $4) bytes
65 * This means $16 is known to be misaligned
66 */
67 EX( ldq_u $5, 0($16) ) # .. .. .. L
68 beq $1, $onebyte # .. .. U ..
69 mskql $5, $16, $5 # .. U .. ..
70 addq $16, 8, $16 # E .. .. ..
71
72 EX( stq_u $5, -8($16) ) # .. .. .. L
73 subq $1, 1, $1 # .. .. E ..
74 addq $0, $4, $0 # .. E .. ..
75 subq $0, 8, $0 # E .. .. ..
76
77 .align 4
78 /*
79 * (The .align directive ought to be a moot po
80 * values upon initial entry to the loop
81 * $1 is number of quadwords to clear (zero is
82 * $2 is number of trailing bytes (0..7) ($2 n
83 * $16 is known to be aligned 0mod8
84 */
85 $headalign:
86 subq $1, 16, $4 # .. .. .. E
87 and $16, 0x3f, $2 # .. .. E ..
88 subq $2, 0x40, $3 # .. E .. ..
89 blt $4, $trailquad # U .. .. ..
90
91 /*
92 * We know that we're going to do at least 16
93 * going to be able to use the large block cle
94 * Figure out how many quads we need to clear
95 * so we can use the wh64 instruction.
96 */
97
98 nop # .. .. .. E
99 nop # .. .. E ..
100 nop # .. E .. ..
101 beq $3, $bigalign # U .. .. ..
102
103 $alignmod64:
104 EX( stq_u $31, 0($16) ) # .. .. .. L
105 addq $3, 8, $3 # .. .. E ..
106 subq $0, 8, $0 # .. E .. ..
107 nop # E .. .. ..
108
109 nop # .. .. .. E
110 subq $1, 1, $1 # .. .. E ..
111 addq $16, 8, $16 # .. E .. ..
112 blt $3, $alignmod64 # U .. .. ..
113
114 $bigalign:
115 /*
116 * $0 is the number of bytes left
117 * $1 is the number of quads left
118 * $16 is aligned 0mod64
119 * we know that we'll be taking a minimum of o
120 * CWG Section 3.7.6: do not expect a sustaine
121 * We are _not_ going to update $0 after every
122 * would be silly, because there will be cross
123 * no matter how the code is scheduled. By do
124 * staggered fashion, we can still do this loo
125 * The worse case will be doing two extra quad
126 * in the event of an interrupted clear.
127 * Assumes the wh64 needs to be for 2 trips th
128 * The wh64 is issued on for the starting dest
129 * through the loop, and if there are less tha
130 * address will be for the current trip.
131 */
132 nop # E :
133 nop # E :
134 nop # E :
135 bis $16,$16,$3 # E : U L U L
136 /* This might actually help for the cu
137
138 $do_wh64:
139 wh64 ($3) # .. .. .. L1
140 subq $1, 16, $4 # .. .. E ..
141 EX( stq_u $31, 0($16) ) # .. L .. ..
142 subq $0, 8, $0 # E .. .. ..
143
144 addq $16, 128, $3 # E : Target a
145 EX( stq_u $31, 8($16) ) # L :
146 EX( stq_u $31, 16($16) ) # L :
147 subq $0, 16, $0 # E : U L L U
148
149 nop # E :
150 EX( stq_u $31, 24($16) ) # L :
151 EX( stq_u $31, 32($16) ) # L :
152 subq $0, 168, $5 # E : U L L U
153 /* 168 = 192 - 24, since we've already
154
155 subq $0, 16, $0 # E :
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
159
160 subq $1, 8, $1 # E :
161 subq $0, 16, $0 # E :
162 EX( stq_u $31, 56($16) ) # L :
163 nop # E : U L U L
164
165 nop # E :
166 subq $0, 8, $0 # E :
167 addq $16, 64, $16 # E :
168 bge $4, $do_wh64 # U : U L U L
169
170 $trailquad:
171 # zero to 16 quadwords left to store,
172 # $1 is the number of quadwords left t
173 #
174 nop # .. .. .. E
175 nop # .. .. E ..
176 nop # .. E .. ..
177 beq $1, $trailbytes # U .. .. ..
178
179 $onequad:
180 EX( stq_u $31, 0($16) ) # .. .. .. L
181 subq $1, 1, $1 # .. .. E ..
182 subq $0, 8, $0 # .. E .. ..
183 nop # E .. .. ..
184
185 nop # .. .. .. E
186 nop # .. .. E ..
187 addq $16, 8, $16 # .. E .. ..
188 bgt $1, $onequad # U .. .. ..
189
190 # We have an unknown number of bytes l
191 $trailbytes:
192 nop # .. .. .. E
193 nop # .. .. E ..
194 nop # .. E .. ..
195 beq $0, $zerolength # U .. .. ..
196
197 # $0 contains the number of bytes left
198 # so we will use $0 as the loop counte
199 # We know for a fact that $0 > 0 zero
200 $onebyte:
201 EX( stb $31, 0($16) ) # .. .. .. L
202 subq $0, 1, $0 # .. .. E ..
203 addq $16, 1, $16 # .. E .. ..
204 bgt $0, $onebyte # U .. .. ..
205
206 $zerolength:
207 $exception: # Destination
208 nop # .. .. .. E
209 nop # .. .. E ..
210 nop # .. E .. ..
211 ret $31, ($26), 1 # L0 .. .. ..
212 .end __clear_user
213 EXPORT_SYMBOL(__clear_user)
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