/* 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 <linux/export.h>
#include <asm/asmmacro.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)