/*
 *
 * 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)