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Diffstat (limited to 'arch/ia64/lib/memcpy.S')
| -rw-r--r-- | arch/ia64/lib/memcpy.S | 301 |
1 files changed, 301 insertions, 0 deletions
diff --git a/arch/ia64/lib/memcpy.S b/arch/ia64/lib/memcpy.S new file mode 100644 index 000000000..448908d80 --- /dev/null +++ b/arch/ia64/lib/memcpy.S @@ -0,0 +1,301 @@ +/* + * + * 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> + +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) |