3 * Optimized version of the standard memcpy() function
6 * in0: destination address
8 * in2: number of bytes to copy
12 * Copyright (C) 2000-2001 Hewlett-Packard Co
13 * Stephane Eranian <eranian@hpl.hp.com>
14 * David Mosberger-Tang <davidm@hpl.hp.com>
16 #include <linux/config.h>
18 #include <asm/asmmacro.h>
26 // gas doesn't handle control flow across procedures, so it doesn't
27 // realize that a stop bit is needed before the "alloc" instruction
38 # define MEM_LAT 21 /* latency to memory */
55 # define N (MEM_LAT + 4)
56 # define Nrot ((N + 7) & ~7)
59 * First, check if everything (src, dst, len) is a multiple of eight. If
60 * so, we handle everything with no taken branches (other than the loop
61 * itself) and a small icache footprint. Otherwise, we jump off to
62 * the more general copy routine handling arbitrary
63 * sizes/alignment etc.
66 .save ar.pfs, saved_pfs
67 alloc saved_pfs=ar.pfs,3,Nrot,0,Nrot
79 cmp.eq p6,p0=in2,r0 // zero length?
80 mov retval=in0 // return dst
81 (p6) br.ret.spnt.many rp // zero length, return immediately
84 mov dst=in0 // copy because of rotation
85 shr.u cnt=in2,3 // number of 8-byte words to copy
89 adds cnt=-1,cnt // br.ctop is repeat/until
90 cmp.gtu p7,p0=16,in2 // copying less than 16 bytes?
99 mov src=in1 // copy because of rotation
100 (p7) br.cond.spnt.few .memcpy_short
101 (p6) br.cond.spnt.few .memcpy_long
114 (p[0]) ld8 val[0]=[src],8
119 (p[N-1])st8 [dst]=val[N-1],8
130 * Small (<16 bytes) unaligned copying is done via a simple byte-at-the-time
131 * copy loop. This performs relatively poorly on Itanium, but it doesn't
132 * get used very often (gcc inlines small copies) and due to atomicity
133 * issues, we want to avoid read-modify-write of entire words.
137 adds cnt=-1,in2 // br.ctop is repeat/until
153 * It is faster to put a stop bit in the loop here because it makes
154 * the pipeline shorter (and latency is what matters on short copies).
158 (p[0]) ld1 val[0]=[src],1
163 (p[MEM_LAT-1])st1 [dst]=val[MEM_LAT-1],1
173 * Large (>= 16 bytes) copying is done in a fancy way. Latency isn't
174 * an overriding concern here, but throughput is. We first do
175 * sub-word copying until the destination is aligned, then we check
176 * if the source is also aligned. If so, we do a simple load/store-loop
177 * until there are less than 8 bytes left over and then we do the tail,
178 * by storing the last few bytes using sub-word copying. If the source
179 * is not aligned, we branch off to the non-congruent loop.
187 * On Itanium, the pipeline itself runs without stalls. However, br.ctop
188 * seems to introduce an unavoidable bubble in the pipeline so the overall
189 * latency is 2 cycles/iteration. This gives us a _copy_ throughput
190 * of 4 byte/cycle. Still not bad.
194 # define N (MEM_LAT + 5) /* number of stages */
195 # define Nrot ((N+1 + 2 + 7) & ~7) /* number of rotating regs */
197 #define LOG_LOOP_SIZE 6
200 alloc t3=ar.pfs,3,Nrot,0,Nrot // resize register frame
201 and t0=-8,src // t0 = src & ~7
202 and t2=7,src // t2 = src & 7
204 ld8 t0=[t0] // t0 = 1st source word
205 adds src2=7,src // src2 = (src + 7)
206 sub t4=r0,dst // t4 = -dst
208 and src2=-8,src2 // src2 = (src + 7) & ~7
209 shl t2=t2,3 // t2 = 8*(src & 7)
210 shl t4=t4,3 // t4 = 8*(dst & 7)
212 ld8 t1=[src2] // t1 = 1st source word if src is 8-byte aligned, 2nd otherwise
213 sub t3=64,t2 // t3 = 64-8*(src & 7)
218 mov pr=t4,0x38 // (p5,p4,p3)=(dst & 7)
222 adds src_end=-1,src_end
234 and src_end=-8,src_end // src_end = last word of source buffer
237 // At this point, dst is aligned to 8 bytes and there at least 16-7=9 bytes left to copy:
239 1:{ add src=cnt,src // make src point to remainder of source buffer
240 sub cnt=in2,cnt // cnt = number of bytes left to copy
243 and src2=-8,src // align source pointer
244 adds t4=.memcpy_loops-1b,t4
247 and t0=7,src // t0 = src & 7
248 shr.u t2=cnt,3 // t2 = number of 8-byte words left to copy
249 shl cnt=cnt,3 // move bits 0-2 to 3-5
255 cmp.ne p6,p0=t0,r0 // is src aligned, too?
256 shl t0=t0,LOG_LOOP_SIZE // t0 = 8*(src & 7)
257 adds t2=-1,t2 // br.ctop is repeat/until
260 mov pr=cnt,0x38 // set (p5,p4,p3) to # of bytes last-word bytes to copy
270 (p6) ld8 val[1]=[src2],8 // prime the pump...
276 // At this point, (p5,p4,p3) are set to the number of bytes left to copy (which is
277 // less than 8) and t0 contains the last few bytes of the src buffer:
290 ///////////////////////////////////////////////////////
293 #define COPY(shift,index) \
295 (p[0]) ld8 val[0]=[src2],8; \
296 (p[MEM_LAT+3]) shrp w[0]=val[MEM_LAT+3],val[MEM_LAT+4-index],shift; \
297 brp.loop.imp 1b, 2f \
300 (p[MEM_LAT+4]) st8 [dst]=w[1],8; \
302 br.ctop.dptk.few 1b; \
305 ld8 val[N-1]=[src_end]; /* load last word (may be same as val[N]) */ \
307 shrp t0=val[N-1],val[N-index],shift; \
310 COPY(0, 1) /* no point special casing this---it doesn't go any faster without shrp */