1 /* SPDX-License-Identifier: GPL-2.0 */
4 * Optimized version of the standard memcpy() function
7 * in0: destination address
9 * in2: number of bytes to copy
13 * Copyright (C) 2000-2001 Hewlett-Packard Co
14 * Stephane Eranian <eranian@hpl.hp.com>
15 * David Mosberger-Tang <davidm@hpl.hp.com>
17 #include <asm/asmmacro.h>
18 #include <asm/export.h>
22 # define MEM_LAT 21 /* latency to memory */
39 # define N (MEM_LAT + 4)
40 # define Nrot ((N + 7) & ~7)
43 * First, check if everything (src, dst, len) is a multiple of eight. If
44 * so, we handle everything with no taken branches (other than the loop
45 * itself) and a small icache footprint. Otherwise, we jump off to
46 * the more general copy routine handling arbitrary
47 * sizes/alignment etc.
50 .save ar.pfs, saved_pfs
51 alloc saved_pfs=ar.pfs,3,Nrot,0,Nrot
63 cmp.eq p6,p0=in2,r0 // zero length?
64 mov retval=in0 // return dst
65 (p6) br.ret.spnt.many rp // zero length, return immediately
68 mov dst=in0 // copy because of rotation
69 shr.u cnt=in2,3 // number of 8-byte words to copy
73 adds cnt=-1,cnt // br.ctop is repeat/until
74 cmp.gtu p7,p0=16,in2 // copying less than 16 bytes?
83 mov src=in1 // copy because of rotation
84 (p7) br.cond.spnt.few .memcpy_short
85 (p6) br.cond.spnt.few .memcpy_long
98 (p[0]) ld8 val[0]=[src],8
103 (p[N-1])st8 [dst]=val[N-1],8
114 * Small (<16 bytes) unaligned copying is done via a simple byte-at-the-time
115 * copy loop. This performs relatively poorly on Itanium, but it doesn't
116 * get used very often (gcc inlines small copies) and due to atomicity
117 * issues, we want to avoid read-modify-write of entire words.
121 adds cnt=-1,in2 // br.ctop is repeat/until
137 * It is faster to put a stop bit in the loop here because it makes
138 * the pipeline shorter (and latency is what matters on short copies).
142 (p[0]) ld1 val[0]=[src],1
147 (p[MEM_LAT-1])st1 [dst]=val[MEM_LAT-1],1
157 * Large (>= 16 bytes) copying is done in a fancy way. Latency isn't
158 * an overriding concern here, but throughput is. We first do
159 * sub-word copying until the destination is aligned, then we check
160 * if the source is also aligned. If so, we do a simple load/store-loop
161 * until there are less than 8 bytes left over and then we do the tail,
162 * by storing the last few bytes using sub-word copying. If the source
163 * is not aligned, we branch off to the non-congruent loop.
171 * On Itanium, the pipeline itself runs without stalls. However, br.ctop
172 * seems to introduce an unavoidable bubble in the pipeline so the overall
173 * latency is 2 cycles/iteration. This gives us a _copy_ throughput
174 * of 4 byte/cycle. Still not bad.
178 # define N (MEM_LAT + 5) /* number of stages */
179 # define Nrot ((N+1 + 2 + 7) & ~7) /* number of rotating regs */
181 #define LOG_LOOP_SIZE 6
184 alloc t3=ar.pfs,3,Nrot,0,Nrot // resize register frame
185 and t0=-8,src // t0 = src & ~7
186 and t2=7,src // t2 = src & 7
188 ld8 t0=[t0] // t0 = 1st source word
189 adds src2=7,src // src2 = (src + 7)
190 sub t4=r0,dst // t4 = -dst
192 and src2=-8,src2 // src2 = (src + 7) & ~7
193 shl t2=t2,3 // t2 = 8*(src & 7)
194 shl t4=t4,3 // t4 = 8*(dst & 7)
196 ld8 t1=[src2] // t1 = 1st source word if src is 8-byte aligned, 2nd otherwise
197 sub t3=64,t2 // t3 = 64-8*(src & 7)
202 mov pr=t4,0x38 // (p5,p4,p3)=(dst & 7)
206 adds src_end=-1,src_end
218 and src_end=-8,src_end // src_end = last word of source buffer
221 // At this point, dst is aligned to 8 bytes and there at least 16-7=9 bytes left to copy:
223 1:{ add src=cnt,src // make src point to remainder of source buffer
224 sub cnt=in2,cnt // cnt = number of bytes left to copy
227 and src2=-8,src // align source pointer
228 adds t4=.memcpy_loops-1b,t4
231 and t0=7,src // t0 = src & 7
232 shr.u t2=cnt,3 // t2 = number of 8-byte words left to copy
233 shl cnt=cnt,3 // move bits 0-2 to 3-5
239 cmp.ne p6,p0=t0,r0 // is src aligned, too?
240 shl t0=t0,LOG_LOOP_SIZE // t0 = 8*(src & 7)
241 adds t2=-1,t2 // br.ctop is repeat/until
244 mov pr=cnt,0x38 // set (p5,p4,p3) to # of bytes last-word bytes to copy
254 (p6) ld8 val[1]=[src2],8 // prime the pump...
260 // At this point, (p5,p4,p3) are set to the number of bytes left to copy (which is
261 // less than 8) and t0 contains the last few bytes of the src buffer:
274 ///////////////////////////////////////////////////////
277 #define COPY(shift,index) \
279 (p[0]) ld8 val[0]=[src2],8; \
280 (p[MEM_LAT+3]) shrp w[0]=val[MEM_LAT+3],val[MEM_LAT+4-index],shift; \
281 brp.loop.imp 1b, 2f \
284 (p[MEM_LAT+4]) st8 [dst]=w[1],8; \
286 br.ctop.dptk.few 1b; \
289 ld8 val[N-1]=[src_end]; /* load last word (may be same as val[N]) */ \
291 shrp t0=val[N-1],val[N-index],shift; \
294 COPY(0, 1) /* no point special casing this---it doesn't go any faster without shrp */
304 EXPORT_SYMBOL(memcpy)