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Linus Torvalds1da177e2005-04-16 15:20:36 -07001/*
2 *
3 * Optimized version of the standard memcpy() function
4 *
5 * Inputs:
6 * in0: destination address
7 * in1: source address
8 * in2: number of bytes to copy
9 * Output:
10 * no return value
11 *
12 * Copyright (C) 2000-2001 Hewlett-Packard Co
13 * Stephane Eranian <eranian@hpl.hp.com>
14 * David Mosberger-Tang <davidm@hpl.hp.com>
15 */
16#include <asm/asmmacro.h>
17
18GLOBAL_ENTRY(memcpy)
19
20# define MEM_LAT 21 /* latency to memory */
21
22# define dst r2
23# define src r3
24# define retval r8
25# define saved_pfs r9
26# define saved_lc r10
27# define saved_pr r11
28# define cnt r16
29# define src2 r17
30# define t0 r18
31# define t1 r19
32# define t2 r20
33# define t3 r21
34# define t4 r22
35# define src_end r23
36
37# define N (MEM_LAT + 4)
38# define Nrot ((N + 7) & ~7)
39
40 /*
41 * First, check if everything (src, dst, len) is a multiple of eight. If
42 * so, we handle everything with no taken branches (other than the loop
43 * itself) and a small icache footprint. Otherwise, we jump off to
44 * the more general copy routine handling arbitrary
45 * sizes/alignment etc.
46 */
47 .prologue
48 .save ar.pfs, saved_pfs
49 alloc saved_pfs=ar.pfs,3,Nrot,0,Nrot
50 .save ar.lc, saved_lc
51 mov saved_lc=ar.lc
52 or t0=in0,in1
53 ;;
54
55 or t0=t0,in2
56 .save pr, saved_pr
57 mov saved_pr=pr
58
59 .body
60
61 cmp.eq p6,p0=in2,r0 // zero length?
62 mov retval=in0 // return dst
63(p6) br.ret.spnt.many rp // zero length, return immediately
64 ;;
65
66 mov dst=in0 // copy because of rotation
67 shr.u cnt=in2,3 // number of 8-byte words to copy
68 mov pr.rot=1<<16
69 ;;
70
71 adds cnt=-1,cnt // br.ctop is repeat/until
72 cmp.gtu p7,p0=16,in2 // copying less than 16 bytes?
73 mov ar.ec=N
74 ;;
75
76 and t0=0x7,t0
77 mov ar.lc=cnt
78 ;;
79 cmp.ne p6,p0=t0,r0
80
81 mov src=in1 // copy because of rotation
82(p7) br.cond.spnt.few .memcpy_short
83(p6) br.cond.spnt.few .memcpy_long
84 ;;
85 nop.m 0
86 ;;
87 nop.m 0
88 nop.i 0
89 ;;
90 nop.m 0
91 ;;
92 .rotr val[N]
93 .rotp p[N]
94 .align 32
951: { .mib
96(p[0]) ld8 val[0]=[src],8
97 nop.i 0
98 brp.loop.imp 1b, 2f
99}
1002: { .mfb
101(p[N-1])st8 [dst]=val[N-1],8
102 nop.f 0
103 br.ctop.dptk.few 1b
104}
105 ;;
106 mov ar.lc=saved_lc
107 mov pr=saved_pr,-1
108 mov ar.pfs=saved_pfs
109 br.ret.sptk.many rp
110
111 /*
112 * Small (<16 bytes) unaligned copying is done via a simple byte-at-the-time
113 * copy loop. This performs relatively poorly on Itanium, but it doesn't
114 * get used very often (gcc inlines small copies) and due to atomicity
115 * issues, we want to avoid read-modify-write of entire words.
116 */
117 .align 32
118.memcpy_short:
119 adds cnt=-1,in2 // br.ctop is repeat/until
120 mov ar.ec=MEM_LAT
121 brp.loop.imp 1f, 2f
122 ;;
123 mov ar.lc=cnt
124 ;;
125 nop.m 0
126 ;;
127 nop.m 0
128 nop.i 0
129 ;;
130 nop.m 0
131 ;;
132 nop.m 0
133 ;;
134 /*
135 * It is faster to put a stop bit in the loop here because it makes
136 * the pipeline shorter (and latency is what matters on short copies).
137 */
138 .align 32
1391: { .mib
140(p[0]) ld1 val[0]=[src],1
141 nop.i 0
142 brp.loop.imp 1b, 2f
143} ;;
1442: { .mfb
145(p[MEM_LAT-1])st1 [dst]=val[MEM_LAT-1],1
146 nop.f 0
147 br.ctop.dptk.few 1b
148} ;;
149 mov ar.lc=saved_lc
150 mov pr=saved_pr,-1
151 mov ar.pfs=saved_pfs
152 br.ret.sptk.many rp
153
154 /*
155 * Large (>= 16 bytes) copying is done in a fancy way. Latency isn't
156 * an overriding concern here, but throughput is. We first do
157 * sub-word copying until the destination is aligned, then we check
158 * if the source is also aligned. If so, we do a simple load/store-loop
159 * until there are less than 8 bytes left over and then we do the tail,
160 * by storing the last few bytes using sub-word copying. If the source
161 * is not aligned, we branch off to the non-congruent loop.
162 *
163 * stage: op:
164 * 0 ld
165 * :
166 * MEM_LAT+3 shrp
167 * MEM_LAT+4 st
168 *
169 * On Itanium, the pipeline itself runs without stalls. However, br.ctop
170 * seems to introduce an unavoidable bubble in the pipeline so the overall
171 * latency is 2 cycles/iteration. This gives us a _copy_ throughput
172 * of 4 byte/cycle. Still not bad.
173 */
174# undef N
175# undef Nrot
176# define N (MEM_LAT + 5) /* number of stages */
177# define Nrot ((N+1 + 2 + 7) & ~7) /* number of rotating regs */
178
179#define LOG_LOOP_SIZE 6
180
181.memcpy_long:
182 alloc t3=ar.pfs,3,Nrot,0,Nrot // resize register frame
183 and t0=-8,src // t0 = src & ~7
184 and t2=7,src // t2 = src & 7
185 ;;
186 ld8 t0=[t0] // t0 = 1st source word
187 adds src2=7,src // src2 = (src + 7)
188 sub t4=r0,dst // t4 = -dst
189 ;;
190 and src2=-8,src2 // src2 = (src + 7) & ~7
191 shl t2=t2,3 // t2 = 8*(src & 7)
192 shl t4=t4,3 // t4 = 8*(dst & 7)
193 ;;
194 ld8 t1=[src2] // t1 = 1st source word if src is 8-byte aligned, 2nd otherwise
195 sub t3=64,t2 // t3 = 64-8*(src & 7)
196 shr.u t0=t0,t2
197 ;;
198 add src_end=src,in2
199 shl t1=t1,t3
200 mov pr=t4,0x38 // (p5,p4,p3)=(dst & 7)
201 ;;
202 or t0=t0,t1
203 mov cnt=r0
204 adds src_end=-1,src_end
205 ;;
206(p3) st1 [dst]=t0,1
207(p3) shr.u t0=t0,8
208(p3) adds cnt=1,cnt
209 ;;
210(p4) st2 [dst]=t0,2
211(p4) shr.u t0=t0,16
212(p4) adds cnt=2,cnt
213 ;;
214(p5) st4 [dst]=t0,4
215(p5) adds cnt=4,cnt
216 and src_end=-8,src_end // src_end = last word of source buffer
217 ;;
218
219 // At this point, dst is aligned to 8 bytes and there at least 16-7=9 bytes left to copy:
220
2211:{ add src=cnt,src // make src point to remainder of source buffer
222 sub cnt=in2,cnt // cnt = number of bytes left to copy
223 mov t4=ip
224 } ;;
225 and src2=-8,src // align source pointer
226 adds t4=.memcpy_loops-1b,t4
227 mov ar.ec=N
228
229 and t0=7,src // t0 = src & 7
230 shr.u t2=cnt,3 // t2 = number of 8-byte words left to copy
231 shl cnt=cnt,3 // move bits 0-2 to 3-5
232 ;;
233
234 .rotr val[N+1], w[2]
235 .rotp p[N]
236
237 cmp.ne p6,p0=t0,r0 // is src aligned, too?
238 shl t0=t0,LOG_LOOP_SIZE // t0 = 8*(src & 7)
239 adds t2=-1,t2 // br.ctop is repeat/until
240 ;;
241 add t4=t0,t4
242 mov pr=cnt,0x38 // set (p5,p4,p3) to # of bytes last-word bytes to copy
243 mov ar.lc=t2
244 ;;
245 nop.m 0
246 ;;
247 nop.m 0
248 nop.i 0
249 ;;
250 nop.m 0
251 ;;
252(p6) ld8 val[1]=[src2],8 // prime the pump...
253 mov b6=t4
254 br.sptk.few b6
255 ;;
256
257.memcpy_tail:
258 // At this point, (p5,p4,p3) are set to the number of bytes left to copy (which is
259 // less than 8) and t0 contains the last few bytes of the src buffer:
260(p5) st4 [dst]=t0,4
261(p5) shr.u t0=t0,32
262 mov ar.lc=saved_lc
263 ;;
264(p4) st2 [dst]=t0,2
265(p4) shr.u t0=t0,16
266 mov ar.pfs=saved_pfs
267 ;;
268(p3) st1 [dst]=t0
269 mov pr=saved_pr,-1
270 br.ret.sptk.many rp
271
272///////////////////////////////////////////////////////
273 .align 64
274
275#define COPY(shift,index) \
276 1: { .mib \
277 (p[0]) ld8 val[0]=[src2],8; \
278 (p[MEM_LAT+3]) shrp w[0]=val[MEM_LAT+3],val[MEM_LAT+4-index],shift; \
279 brp.loop.imp 1b, 2f \
280 }; \
281 2: { .mfb \
282 (p[MEM_LAT+4]) st8 [dst]=w[1],8; \
283 nop.f 0; \
284 br.ctop.dptk.few 1b; \
285 }; \
286 ;; \
287 ld8 val[N-1]=[src_end]; /* load last word (may be same as val[N]) */ \
288 ;; \
289 shrp t0=val[N-1],val[N-index],shift; \
290 br .memcpy_tail
291.memcpy_loops:
292 COPY(0, 1) /* no point special casing this---it doesn't go any faster without shrp */
293 COPY(8, 0)
294 COPY(16, 0)
295 COPY(24, 0)
296 COPY(32, 0)
297 COPY(40, 0)
298 COPY(48, 0)
299 COPY(56, 0)
300
301END(memcpy)