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