Merge branch 'master' of /home/tglx/work/mtd/git/linux-2.6.git/

[linux-2.6] / arch / ia64 / lib / copy_user.S

/*
 *
 * Optimized version of the copy_user() routine.
 * It is used to copy date across the kernel/user boundary.
 *
 * The source and destination are always on opposite side of
 * the boundary. When reading from user space we must catch
 * faults on loads. When writing to user space we must catch
 * errors on stores. Note that because of the nature of the copy
 * we don't need to worry about overlapping regions.
 *
 *
 * Inputs:
 *      in0     address of source buffer
 *      in1     address of destination buffer
 *      in2     number of bytes to copy
 *
 * Outputs:
 *      ret0    0 in case of success. The number of bytes NOT copied in
 *              case of error.
 *
 * Copyright (C) 2000-2001 Hewlett-Packard Co
 *      Stephane Eranian <eranian@hpl.hp.com>
 *
 * Fixme:
 *      - handle the case where we have more than 16 bytes and the alignment
 *        are different.
 *      - more benchmarking
 *      - fix extraneous stop bit introduced by the EX() macro.
 */

#include <asm/asmmacro.h>

//
// Tuneable parameters
//
#define COPY_BREAK      16      // we do byte copy below (must be >=16)
#define PIPE_DEPTH      21      // pipe depth

#define EPI             p[PIPE_DEPTH-1]

//
// arguments
//
#define dst             in0
#define src             in1
#define len             in2

//
// local registers
//
#define t1              r2      // rshift in bytes
#define t2              r3      // lshift in bytes
#define rshift          r14     // right shift in bits
#define lshift          r15     // left shift in bits
#define word1           r16
#define word2           r17
#define cnt             r18
#define len2            r19
#define saved_lc        r20
#define saved_pr        r21
#define tmp             r22
#define val             r23
#define src1            r24
#define dst1            r25
#define src2            r26
#define dst2            r27
#define len1            r28
#define enddst          r29
#define endsrc          r30
#define saved_pfs       r31

GLOBAL_ENTRY(__copy_user)
        .prologue
        .save ar.pfs, saved_pfs
        alloc saved_pfs=ar.pfs,3,((2*PIPE_DEPTH+7)&~7),0,((2*PIPE_DEPTH+7)&~7)

        .rotr val1[PIPE_DEPTH],val2[PIPE_DEPTH]
        .rotp p[PIPE_DEPTH]

        adds len2=-1,len        // br.ctop is repeat/until
        mov ret0=r0

        ;;                      // RAW of cfm when len=0
        cmp.eq p8,p0=r0,len     // check for zero length
        .save ar.lc, saved_lc
        mov saved_lc=ar.lc      // preserve ar.lc (slow)
(p8)    br.ret.spnt.many rp     // empty mempcy()
        ;;
        add enddst=dst,len      // first byte after end of source
        add endsrc=src,len      // first byte after end of destination
        .save pr, saved_pr
        mov saved_pr=pr         // preserve predicates

        .body

        mov dst1=dst            // copy because of rotation
        mov ar.ec=PIPE_DEPTH
        mov pr.rot=1<<16        // p16=true all others are false

        mov src1=src            // copy because of rotation
        mov ar.lc=len2          // initialize lc for small count
        cmp.lt p10,p7=COPY_BREAK,len    // if len > COPY_BREAK then long copy

        xor tmp=src,dst         // same alignment test prepare
(p10)   br.cond.dptk .long_copy_user
        ;;                      // RAW pr.rot/p16 ?
        //
        // Now we do the byte by byte loop with software pipeline
        //
        // p7 is necessarily false by now
1:
        EX(.failure_in_pipe1,(p16) ld1 val1[0]=[src1],1)
        EX(.failure_out,(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1)
        br.ctop.dptk.few 1b
        ;;
        mov ar.lc=saved_lc
        mov pr=saved_pr,0xffffffffffff0000
        mov ar.pfs=saved_pfs            // restore ar.ec
        br.ret.sptk.many rp             // end of short memcpy

        //
        // Not 8-byte aligned
        //
.diff_align_copy_user:
        // At this point we know we have more than 16 bytes to copy
        // and also that src and dest do _not_ have the same alignment.
        and src2=0x7,src1                               // src offset
        and dst2=0x7,dst1                               // dst offset
        ;;
        // The basic idea is that we copy byte-by-byte at the head so
        // that we can reach 8-byte alignment for both src1 and dst1.
        // Then copy the body using software pipelined 8-byte copy,
        // shifting the two back-to-back words right and left, then copy
        // the tail by copying byte-by-byte.
        //
        // Fault handling. If the byte-by-byte at the head fails on the
        // load, then restart and finish the pipleline by copying zeros
        // to the dst1. Then copy zeros for the rest of dst1.
        // If 8-byte software pipeline fails on the load, do the same as
        // failure_in3 does. If the byte-by-byte at the tail fails, it is
        // handled simply by failure_in_pipe1.
        //
        // The case p14 represents the source has more bytes in the
        // the first word (by the shifted part), whereas the p15 needs to
        // copy some bytes from the 2nd word of the source that has the
        // tail of the 1st of the destination.
        //

        //
        // Optimization. If dst1 is 8-byte aligned (quite common), we don't need
        // to copy the head to dst1, to start 8-byte copy software pipeline.
        // We know src1 is not 8-byte aligned in this case.
        //
        cmp.eq p14,p15=r0,dst2
(p15)   br.cond.spnt 1f
        ;;
        sub t1=8,src2
        mov t2=src2
        ;;
        shl rshift=t2,3
        sub len1=len,t1                                 // set len1
        ;;
        sub lshift=64,rshift
        ;;
        br.cond.spnt .word_copy_user
        ;;
1:
        cmp.leu p14,p15=src2,dst2
        sub t1=dst2,src2
        ;;
        .pred.rel "mutex", p14, p15
(p14)   sub word1=8,src2                                // (8 - src offset)
(p15)   sub t1=r0,t1                                    // absolute value
(p15)   sub word1=8,dst2                                // (8 - dst offset)
        ;;
        // For the case p14, we don't need to copy the shifted part to
        // the 1st word of destination.
        sub t2=8,t1
(p14)   sub word1=word1,t1
        ;;
        sub len1=len,word1                              // resulting len
(p15)   shl rshift=t1,3                                 // in bits
(p14)   shl rshift=t2,3
        ;;
(p14)   sub len1=len1,t1
        adds cnt=-1,word1
        ;;
        sub lshift=64,rshift
        mov ar.ec=PIPE_DEPTH
        mov pr.rot=1<<16        // p16=true all others are false
        mov ar.lc=cnt
        ;;
2:
        EX(.failure_in_pipe2,(p16) ld1 val1[0]=[src1],1)
        EX(.failure_out,(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1)
        br.ctop.dptk.few 2b
        ;;
        clrrrb
        ;;
.word_copy_user:
        cmp.gtu p9,p0=16,len1
(p9)    br.cond.spnt 4f                 // if (16 > len1) skip 8-byte copy
        ;;
        shr.u cnt=len1,3                // number of 64-bit words
        ;;
        adds cnt=-1,cnt
        ;;
        .pred.rel "mutex", p14, p15
(p14)   sub src1=src1,t2
(p15)   sub src1=src1,t1
        //
        // Now both src1 and dst1 point to an 8-byte aligned address. And
        // we have more than 8 bytes to copy.
        //
        mov ar.lc=cnt
        mov ar.ec=PIPE_DEPTH
        mov pr.rot=1<<16        // p16=true all others are false
        ;;
3:
        //
        // The pipleline consists of 3 stages:
        // 1 (p16):     Load a word from src1
        // 2 (EPI_1):   Shift right pair, saving to tmp
        // 3 (EPI):     Store tmp to dst1
        //
        // To make it simple, use at least 2 (p16) loops to set up val1[n]
        // because we need 2 back-to-back val1[] to get tmp.
        // Note that this implies EPI_2 must be p18 or greater.
        //

#define EPI_1           p[PIPE_DEPTH-2]
#define SWITCH(pred, shift)     cmp.eq pred,p0=shift,rshift
#define CASE(pred, shift)       \
        (pred)  br.cond.spnt .copy_user_bit##shift
#define BODY(rshift)                                            \
.copy_user_bit##rshift:                                         \
1:                                                              \
        EX(.failure_out,(EPI) st8 [dst1]=tmp,8);                \
(EPI_1) shrp tmp=val1[PIPE_DEPTH-2],val1[PIPE_DEPTH-1],rshift;  \
        EX(3f,(p16) ld8 val1[1]=[src1],8);                      \
(p16)   mov val1[0]=r0;                                         \
        br.ctop.dptk 1b;                                        \
        ;;                                                      \
        br.cond.sptk.many .diff_align_do_tail;                  \
2:                                                              \
(EPI)   st8 [dst1]=tmp,8;                                       \
(EPI_1) shrp tmp=val1[PIPE_DEPTH-2],val1[PIPE_DEPTH-1],rshift;  \
3:                                                              \
(p16)   mov val1[1]=r0;                                         \
(p16)   mov val1[0]=r0;                                         \
        br.ctop.dptk 2b;                                        \
        ;;                                                      \
        br.cond.sptk.many .failure_in2

        //
        // Since the instruction 'shrp' requires a fixed 128-bit value
        // specifying the bits to shift, we need to provide 7 cases
        // below.
        //
        SWITCH(p6, 8)
        SWITCH(p7, 16)
        SWITCH(p8, 24)
        SWITCH(p9, 32)
        SWITCH(p10, 40)
        SWITCH(p11, 48)
        SWITCH(p12, 56)
        ;;
        CASE(p6, 8)
        CASE(p7, 16)
        CASE(p8, 24)
        CASE(p9, 32)
        CASE(p10, 40)
        CASE(p11, 48)
        CASE(p12, 56)
        ;;
        BODY(8)
        BODY(16)
        BODY(24)
        BODY(32)
        BODY(40)
        BODY(48)
        BODY(56)
        ;;
.diff_align_do_tail:
        .pred.rel "mutex", p14, p15
(p14)   sub src1=src1,t1
(p14)   adds dst1=-8,dst1
(p15)   sub dst1=dst1,t1
        ;;
4:
        // Tail correction.
        //
        // The problem with this piplelined loop is that the last word is not
        // loaded and thus parf of the last word written is not correct.
        // To fix that, we simply copy the tail byte by byte.

        sub len1=endsrc,src1,1
        clrrrb
        ;;
        mov ar.ec=PIPE_DEPTH
        mov pr.rot=1<<16        // p16=true all others are false
        mov ar.lc=len1
        ;;
5:
        EX(.failure_in_pipe1,(p16) ld1 val1[0]=[src1],1)
        EX(.failure_out,(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1)
        br.ctop.dptk.few 5b
        ;;
        mov ar.lc=saved_lc
        mov pr=saved_pr,0xffffffffffff0000
        mov ar.pfs=saved_pfs
        br.ret.sptk.many rp

        //
        // Beginning of long mempcy (i.e. > 16 bytes)
        //
.long_copy_user:
        tbit.nz p6,p7=src1,0    // odd alignment
        and tmp=7,tmp
        ;;
        cmp.eq p10,p8=r0,tmp
        mov len1=len            // copy because of rotation
(p8)    br.cond.dpnt .diff_align_copy_user
        ;;
        // At this point we know we have more than 16 bytes to copy
        // and also that both src and dest have the same alignment
        // which may not be the one we want. So for now we must move
        // forward slowly until we reach 16byte alignment: no need to
        // worry about reaching the end of buffer.
        //
        EX(.failure_in1,(p6) ld1 val1[0]=[src1],1)      // 1-byte aligned
(p6)    adds len1=-1,len1;;
        tbit.nz p7,p0=src1,1
        ;;
        EX(.failure_in1,(p7) ld2 val1[1]=[src1],2)      // 2-byte aligned
(p7)    adds len1=-2,len1;;
        tbit.nz p8,p0=src1,2
        ;;
        //
        // Stop bit not required after ld4 because if we fail on ld4
        // we have never executed the ld1, therefore st1 is not executed.
        //
        EX(.failure_in1,(p8) ld4 val2[0]=[src1],4)      // 4-byte aligned
        ;;
        EX(.failure_out,(p6) st1 [dst1]=val1[0],1)
        tbit.nz p9,p0=src1,3
        ;;
        //
        // Stop bit not required after ld8 because if we fail on ld8
        // we have never executed the ld2, therefore st2 is not executed.
        //
        EX(.failure_in1,(p9) ld8 val2[1]=[src1],8)      // 8-byte aligned
        EX(.failure_out,(p7) st2 [dst1]=val1[1],2)
(p8)    adds len1=-4,len1
        ;;
        EX(.failure_out, (p8) st4 [dst1]=val2[0],4)
(p9)    adds len1=-8,len1;;
        shr.u cnt=len1,4                // number of 128-bit (2x64bit) words
        ;;
        EX(.failure_out, (p9) st8 [dst1]=val2[1],8)
        tbit.nz p6,p0=len1,3
        cmp.eq p7,p0=r0,cnt
        adds tmp=-1,cnt                 // br.ctop is repeat/until
(p7)    br.cond.dpnt .dotail            // we have less than 16 bytes left
        ;;
        adds src2=8,src1
        adds dst2=8,dst1
        mov ar.lc=tmp
        ;;
        //
        // 16bytes/iteration
        //
2:
        EX(.failure_in3,(p16) ld8 val1[0]=[src1],16)
(p16)   ld8 val2[0]=[src2],16

        EX(.failure_out, (EPI)  st8 [dst1]=val1[PIPE_DEPTH-1],16)
(EPI)   st8 [dst2]=val2[PIPE_DEPTH-1],16
        br.ctop.dptk 2b
        ;;                      // RAW on src1 when fall through from loop
        //
        // Tail correction based on len only
        //
        // No matter where we come from (loop or test) the src1 pointer
        // is 16 byte aligned AND we have less than 16 bytes to copy.
        //
.dotail:
        EX(.failure_in1,(p6) ld8 val1[0]=[src1],8)      // at least 8 bytes
        tbit.nz p7,p0=len1,2
        ;;
        EX(.failure_in1,(p7) ld4 val1[1]=[src1],4)      // at least 4 bytes
        tbit.nz p8,p0=len1,1
        ;;
        EX(.failure_in1,(p8) ld2 val2[0]=[src1],2)      // at least 2 bytes
        tbit.nz p9,p0=len1,0
        ;;
        EX(.failure_out, (p6) st8 [dst1]=val1[0],8)
        ;;
        EX(.failure_in1,(p9) ld1 val2[1]=[src1])        // only 1 byte left
        mov ar.lc=saved_lc
        ;;
        EX(.failure_out,(p7) st4 [dst1]=val1[1],4)
        mov pr=saved_pr,0xffffffffffff0000
        ;;
        EX(.failure_out, (p8)   st2 [dst1]=val2[0],2)
        mov ar.pfs=saved_pfs
        ;;
        EX(.failure_out, (p9)   st1 [dst1]=val2[1])
        br.ret.sptk.many rp


        //
        // Here we handle the case where the byte by byte copy fails
        // on the load.
        // Several factors make the zeroing of the rest of the buffer kind of
        // tricky:
        //      - the pipeline: loads/stores are not in sync (pipeline)
        //
        //        In the same loop iteration, the dst1 pointer does not directly
        //        reflect where the faulty load was.
        //
        //      - pipeline effect
        //        When you get a fault on load, you may have valid data from
        //        previous loads not yet store in transit. Such data must be
        //        store normally before moving onto zeroing the rest.
        //
        //      - single/multi dispersal independence.
        //
        // solution:
        //      - we don't disrupt the pipeline, i.e. data in transit in
        //        the software pipeline will be eventually move to memory.
        //        We simply replace the load with a simple mov and keep the
        //        pipeline going. We can't really do this inline because
        //        p16 is always reset to 1 when lc > 0.
        //
.failure_in_pipe1:
        sub ret0=endsrc,src1    // number of bytes to zero, i.e. not copied
1:
(p16)   mov val1[0]=r0
(EPI)   st1 [dst1]=val1[PIPE_DEPTH-1],1
        br.ctop.dptk 1b
        ;;
        mov pr=saved_pr,0xffffffffffff0000
        mov ar.lc=saved_lc
        mov ar.pfs=saved_pfs
        br.ret.sptk.many rp

        //
        // This is the case where the byte by byte copy fails on the load
        // when we copy the head. We need to finish the pipeline and copy
        // zeros for the rest of the destination. Since this happens
        // at the top we still need to fill the body and tail.
.failure_in_pipe2:
        sub ret0=endsrc,src1    // number of bytes to zero, i.e. not copied
2:
(p16)   mov val1[0]=r0
(EPI)   st1 [dst1]=val1[PIPE_DEPTH-1],1
        br.ctop.dptk 2b
        ;;
        sub len=enddst,dst1,1           // precompute len
        br.cond.dptk.many .failure_in1bis
        ;;

        //
        // Here we handle the head & tail part when we check for alignment.
        // The following code handles only the load failures. The
        // main diffculty comes from the fact that loads/stores are
        // scheduled. So when you fail on a load, the stores corresponding
        // to previous successful loads must be executed.
        //
        // However some simplifications are possible given the way
        // things work.
        //
        // 1) HEAD
        // Theory of operation:
        //
        //  Page A   | Page B
        //  ---------|-----
        //          1|8 x
        //        1 2|8 x
        //          4|8 x
        //        1 4|8 x
        //        2 4|8 x
        //      1 2 4|8 x
        //           |1
        //           |2 x
        //           |4 x
        //
        // page_size >= 4k (2^12).  (x means 4, 2, 1)
        // Here we suppose Page A exists and Page B does not.
        //
        // As we move towards eight byte alignment we may encounter faults.
        // The numbers on each page show the size of the load (current alignment).
        //
        // Key point:
        //      - if you fail on 1, 2, 4 then you have never executed any smaller
        //        size loads, e.g. failing ld4 means no ld1 nor ld2 executed
        //        before.
        //
        // This allows us to simplify the cleanup code, because basically you
        // only have to worry about "pending" stores in the case of a failing
        // ld8(). Given the way the code is written today, this means only
        // worry about st2, st4. There we can use the information encapsulated
        // into the predicates.
        //
        // Other key point:
        //      - if you fail on the ld8 in the head, it means you went straight
        //        to it, i.e. 8byte alignment within an unexisting page.
        // Again this comes from the fact that if you crossed just for the ld8 then
        // you are 8byte aligned but also 16byte align, therefore you would
        // either go for the 16byte copy loop OR the ld8 in the tail part.
        // The combination ld1, ld2, ld4, ld8 where you fail on ld8 is impossible
        // because it would mean you had 15bytes to copy in which case you
        // would have defaulted to the byte by byte copy.
        //
        //
        // 2) TAIL
        // Here we now we have less than 16 bytes AND we are either 8 or 16 byte
        // aligned.
        //
        // Key point:
        // This means that we either:
        //              - are right on a page boundary
        //      OR
        //              - are at more than 16 bytes from a page boundary with
        //                at most 15 bytes to copy: no chance of crossing.
        //
        // This allows us to assume that if we fail on a load we haven't possibly
        // executed any of the previous (tail) ones, so we don't need to do
        // any stores. For instance, if we fail on ld2, this means we had
        // 2 or 3 bytes left to copy and we did not execute the ld8 nor ld4.
        //
        // This means that we are in a situation similar the a fault in the
        // head part. That's nice!
        //
.failure_in1:
        sub ret0=endsrc,src1    // number of bytes to zero, i.e. not copied
        sub len=endsrc,src1,1
        //
        // we know that ret0 can never be zero at this point
        // because we failed why trying to do a load, i.e. there is still
        // some work to do.
        // The failure_in1bis and length problem is taken care of at the
        // calling side.
        //
        ;;
.failure_in1bis:                // from (.failure_in3)
        mov ar.lc=len           // Continue with a stupid byte store.
        ;;
5:
        st1 [dst1]=r0,1
        br.cloop.dptk 5b
        ;;
        mov pr=saved_pr,0xffffffffffff0000
        mov ar.lc=saved_lc
        mov ar.pfs=saved_pfs
        br.ret.sptk.many rp

        //
        // Here we simply restart the loop but instead
        // of doing loads we fill the pipeline with zeroes
        // We can't simply store r0 because we may have valid
        // data in transit in the pipeline.
        // ar.lc and ar.ec are setup correctly at this point
        //
        // we MUST use src1/endsrc here and not dst1/enddst because
        // of the pipeline effect.
        //
.failure_in3:
        sub ret0=endsrc,src1    // number of bytes to zero, i.e. not copied
        ;;
2:
(p16)   mov val1[0]=r0
(p16)   mov val2[0]=r0
(EPI)   st8 [dst1]=val1[PIPE_DEPTH-1],16
(EPI)   st8 [dst2]=val2[PIPE_DEPTH-1],16
        br.ctop.dptk 2b
        ;;
        cmp.ne p6,p0=dst1,enddst        // Do we need to finish the tail ?
        sub len=enddst,dst1,1           // precompute len
(p6)    br.cond.dptk .failure_in1bis
        ;;
        mov pr=saved_pr,0xffffffffffff0000
        mov ar.lc=saved_lc
        mov ar.pfs=saved_pfs
        br.ret.sptk.many rp

.failure_in2:
        sub ret0=endsrc,src1
        cmp.ne p6,p0=dst1,enddst        // Do we need to finish the tail ?
        sub len=enddst,dst1,1           // precompute len
(p6)    br.cond.dptk .failure_in1bis
        ;;
        mov pr=saved_pr,0xffffffffffff0000
        mov ar.lc=saved_lc
        mov ar.pfs=saved_pfs
        br.ret.sptk.many rp

        //
        // handling of failures on stores: that's the easy part
        //
.failure_out:
        sub ret0=enddst,dst1
        mov pr=saved_pr,0xffffffffffff0000
        mov ar.lc=saved_lc

        mov ar.pfs=saved_pfs
        br.ret.sptk.many rp
END(__copy_user)