| /* SPDX-License-Identifier: GPL-2.0 */ |
| /* |
| * |
| * Optimized version of the standard memcpy() function |
| * |
| * Inputs: |
| * in0: destination address |
| * in1: source address |
| * in2: number of bytes to copy |
| * Output: |
| * no return value |
| * |
| * Copyright (C) 2000-2001 Hewlett-Packard Co |
| * Stephane Eranian <eranian@hpl.hp.com> |
| * David Mosberger-Tang <davidm@hpl.hp.com> |
| */ |
| #include <asm/asmmacro.h> |
| #include <asm/export.h> |
| |
| GLOBAL_ENTRY(memcpy) |
| |
| # define MEM_LAT 21 /* latency to memory */ |
| |
| # define dst r2 |
| # define src r3 |
| # define retval r8 |
| # define saved_pfs r9 |
| # define saved_lc r10 |
| # define saved_pr r11 |
| # define cnt r16 |
| # define src2 r17 |
| # define t0 r18 |
| # define t1 r19 |
| # define t2 r20 |
| # define t3 r21 |
| # define t4 r22 |
| # define src_end r23 |
| |
| # define N (MEM_LAT + 4) |
| # define Nrot ((N + 7) & ~7) |
| |
| /* |
| * First, check if everything (src, dst, len) is a multiple of eight. If |
| * so, we handle everything with no taken branches (other than the loop |
| * itself) and a small icache footprint. Otherwise, we jump off to |
| * the more general copy routine handling arbitrary |
| * sizes/alignment etc. |
| */ |
| .prologue |
| .save ar.pfs, saved_pfs |
| alloc saved_pfs=ar.pfs,3,Nrot,0,Nrot |
| .save ar.lc, saved_lc |
| mov saved_lc=ar.lc |
| or t0=in0,in1 |
| ;; |
| |
| or t0=t0,in2 |
| .save pr, saved_pr |
| mov saved_pr=pr |
| |
| .body |
| |
| cmp.eq p6,p0=in2,r0 // zero length? |
| mov retval=in0 // return dst |
| (p6) br.ret.spnt.many rp // zero length, return immediately |
| ;; |
| |
| mov dst=in0 // copy because of rotation |
| shr.u cnt=in2,3 // number of 8-byte words to copy |
| mov pr.rot=1<<16 |
| ;; |
| |
| adds cnt=-1,cnt // br.ctop is repeat/until |
| cmp.gtu p7,p0=16,in2 // copying less than 16 bytes? |
| mov ar.ec=N |
| ;; |
| |
| and t0=0x7,t0 |
| mov ar.lc=cnt |
| ;; |
| cmp.ne p6,p0=t0,r0 |
| |
| mov src=in1 // copy because of rotation |
| (p7) br.cond.spnt.few .memcpy_short |
| (p6) br.cond.spnt.few .memcpy_long |
| ;; |
| nop.m 0 |
| ;; |
| nop.m 0 |
| nop.i 0 |
| ;; |
| nop.m 0 |
| ;; |
| .rotr val[N] |
| .rotp p[N] |
| .align 32 |
| 1: { .mib |
| (p[0]) ld8 val[0]=[src],8 |
| nop.i 0 |
| brp.loop.imp 1b, 2f |
| } |
| 2: { .mfb |
| (p[N-1])st8 [dst]=val[N-1],8 |
| nop.f 0 |
| br.ctop.dptk.few 1b |
| } |
| ;; |
| mov ar.lc=saved_lc |
| mov pr=saved_pr,-1 |
| mov ar.pfs=saved_pfs |
| br.ret.sptk.many rp |
| |
| /* |
| * Small (<16 bytes) unaligned copying is done via a simple byte-at-the-time |
| * copy loop. This performs relatively poorly on Itanium, but it doesn't |
| * get used very often (gcc inlines small copies) and due to atomicity |
| * issues, we want to avoid read-modify-write of entire words. |
| */ |
| .align 32 |
| .memcpy_short: |
| adds cnt=-1,in2 // br.ctop is repeat/until |
| mov ar.ec=MEM_LAT |
| brp.loop.imp 1f, 2f |
| ;; |
| mov ar.lc=cnt |
| ;; |
| nop.m 0 |
| ;; |
| nop.m 0 |
| nop.i 0 |
| ;; |
| nop.m 0 |
| ;; |
| nop.m 0 |
| ;; |
| /* |
| * It is faster to put a stop bit in the loop here because it makes |
| * the pipeline shorter (and latency is what matters on short copies). |
| */ |
| .align 32 |
| 1: { .mib |
| (p[0]) ld1 val[0]=[src],1 |
| nop.i 0 |
| brp.loop.imp 1b, 2f |
| } ;; |
| 2: { .mfb |
| (p[MEM_LAT-1])st1 [dst]=val[MEM_LAT-1],1 |
| nop.f 0 |
| br.ctop.dptk.few 1b |
| } ;; |
| mov ar.lc=saved_lc |
| mov pr=saved_pr,-1 |
| mov ar.pfs=saved_pfs |
| br.ret.sptk.many rp |
| |
| /* |
| * Large (>= 16 bytes) copying is done in a fancy way. Latency isn't |
| * an overriding concern here, but throughput is. We first do |
| * sub-word copying until the destination is aligned, then we check |
| * if the source is also aligned. If so, we do a simple load/store-loop |
| * until there are less than 8 bytes left over and then we do the tail, |
| * by storing the last few bytes using sub-word copying. If the source |
| * is not aligned, we branch off to the non-congruent loop. |
| * |
| * stage: op: |
| * 0 ld |
| * : |
| * MEM_LAT+3 shrp |
| * MEM_LAT+4 st |
| * |
| * On Itanium, the pipeline itself runs without stalls. However, br.ctop |
| * seems to introduce an unavoidable bubble in the pipeline so the overall |
| * latency is 2 cycles/iteration. This gives us a _copy_ throughput |
| * of 4 byte/cycle. Still not bad. |
| */ |
| # undef N |
| # undef Nrot |
| # define N (MEM_LAT + 5) /* number of stages */ |
| # define Nrot ((N+1 + 2 + 7) & ~7) /* number of rotating regs */ |
| |
| #define LOG_LOOP_SIZE 6 |
| |
| .memcpy_long: |
| alloc t3=ar.pfs,3,Nrot,0,Nrot // resize register frame |
| and t0=-8,src // t0 = src & ~7 |
| and t2=7,src // t2 = src & 7 |
| ;; |
| ld8 t0=[t0] // t0 = 1st source word |
| adds src2=7,src // src2 = (src + 7) |
| sub t4=r0,dst // t4 = -dst |
| ;; |
| and src2=-8,src2 // src2 = (src + 7) & ~7 |
| shl t2=t2,3 // t2 = 8*(src & 7) |
| shl t4=t4,3 // t4 = 8*(dst & 7) |
| ;; |
| ld8 t1=[src2] // t1 = 1st source word if src is 8-byte aligned, 2nd otherwise |
| sub t3=64,t2 // t3 = 64-8*(src & 7) |
| shr.u t0=t0,t2 |
| ;; |
| add src_end=src,in2 |
| shl t1=t1,t3 |
| mov pr=t4,0x38 // (p5,p4,p3)=(dst & 7) |
| ;; |
| or t0=t0,t1 |
| mov cnt=r0 |
| adds src_end=-1,src_end |
| ;; |
| (p3) st1 [dst]=t0,1 |
| (p3) shr.u t0=t0,8 |
| (p3) adds cnt=1,cnt |
| ;; |
| (p4) st2 [dst]=t0,2 |
| (p4) shr.u t0=t0,16 |
| (p4) adds cnt=2,cnt |
| ;; |
| (p5) st4 [dst]=t0,4 |
| (p5) adds cnt=4,cnt |
| and src_end=-8,src_end // src_end = last word of source buffer |
| ;; |
| |
| // At this point, dst is aligned to 8 bytes and there at least 16-7=9 bytes left to copy: |
| |
| 1:{ add src=cnt,src // make src point to remainder of source buffer |
| sub cnt=in2,cnt // cnt = number of bytes left to copy |
| mov t4=ip |
| } ;; |
| and src2=-8,src // align source pointer |
| adds t4=.memcpy_loops-1b,t4 |
| mov ar.ec=N |
| |
| and t0=7,src // t0 = src & 7 |
| shr.u t2=cnt,3 // t2 = number of 8-byte words left to copy |
| shl cnt=cnt,3 // move bits 0-2 to 3-5 |
| ;; |
| |
| .rotr val[N+1], w[2] |
| .rotp p[N] |
| |
| cmp.ne p6,p0=t0,r0 // is src aligned, too? |
| shl t0=t0,LOG_LOOP_SIZE // t0 = 8*(src & 7) |
| adds t2=-1,t2 // br.ctop is repeat/until |
| ;; |
| add t4=t0,t4 |
| mov pr=cnt,0x38 // set (p5,p4,p3) to # of bytes last-word bytes to copy |
| mov ar.lc=t2 |
| ;; |
| nop.m 0 |
| ;; |
| nop.m 0 |
| nop.i 0 |
| ;; |
| nop.m 0 |
| ;; |
| (p6) ld8 val[1]=[src2],8 // prime the pump... |
| mov b6=t4 |
| br.sptk.few b6 |
| ;; |
| |
| .memcpy_tail: |
| // At this point, (p5,p4,p3) are set to the number of bytes left to copy (which is |
| // less than 8) and t0 contains the last few bytes of the src buffer: |
| (p5) st4 [dst]=t0,4 |
| (p5) shr.u t0=t0,32 |
| mov ar.lc=saved_lc |
| ;; |
| (p4) st2 [dst]=t0,2 |
| (p4) shr.u t0=t0,16 |
| mov ar.pfs=saved_pfs |
| ;; |
| (p3) st1 [dst]=t0 |
| mov pr=saved_pr,-1 |
| br.ret.sptk.many rp |
| |
| /////////////////////////////////////////////////////// |
| .align 64 |
| |
| #define COPY(shift,index) \ |
| 1: { .mib \ |
| (p[0]) ld8 val[0]=[src2],8; \ |
| (p[MEM_LAT+3]) shrp w[0]=val[MEM_LAT+3],val[MEM_LAT+4-index],shift; \ |
| brp.loop.imp 1b, 2f \ |
| }; \ |
| 2: { .mfb \ |
| (p[MEM_LAT+4]) st8 [dst]=w[1],8; \ |
| nop.f 0; \ |
| br.ctop.dptk.few 1b; \ |
| }; \ |
| ;; \ |
| ld8 val[N-1]=[src_end]; /* load last word (may be same as val[N]) */ \ |
| ;; \ |
| shrp t0=val[N-1],val[N-index],shift; \ |
| br .memcpy_tail |
| .memcpy_loops: |
| COPY(0, 1) /* no point special casing this---it doesn't go any faster without shrp */ |
| COPY(8, 0) |
| COPY(16, 0) |
| COPY(24, 0) |
| COPY(32, 0) |
| COPY(40, 0) |
| COPY(48, 0) |
| COPY(56, 0) |
| |
| END(memcpy) |
| EXPORT_SYMBOL(memcpy) |