arch/ia64/lib/memcpy.S - arm/linux - Git at Google

 /*
  *
  * 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)
	/*
	*
	* 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)