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/*
 * arch/alpha/lib/ev6-copy_user.S
 *
 * 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com>
 *
 * Copy to/from user space, handling exceptions as we go..  This
 * isn't exactly pretty.
 *
 * This is essentially the same as "memcpy()", but with a few twists.
 * Notably, we have to make sure that $0 is always up-to-date and
 * contains the right "bytes left to copy" value (and that it is updated
 * only _after_ a successful copy). There is also some rather minor
 * exception setup stuff..
 *
 * NOTE! This is not directly C-callable, because the calling semantics are
 * different:
 *
 * Inputs:
 *      length in $0
 *      destination address in $6
 *      source address in $7
 *      return address in $28
 *
 * Outputs:
 *      bytes left to copy in $0
 *
 * Clobbers:
 *      $1,$2,$3,$4,$5,$6,$7
 *
 * Much of the information about 21264 scheduling/coding comes from:
 *      Compiler Writer's Guide for the Alpha 21264
 *      abbreviated as 'CWG' in other comments here
 *      ftp.digital.com/pub/Digital/info/semiconductor/literature/dsc-library.html
 * Scheduling notation:
 *      E       - either cluster
 *      U       - upper subcluster; U0 - subcluster U0; U1 - subcluster U1
 *      L       - lower subcluster; L0 - subcluster L0; L1 - subcluster L1
 */

/* Allow an exception for an insn; exit if we get one.  */
#define EXI(x,y...)                     \
        99: x,##y;                      \
        .section __ex_table,"a";        \
        .gprel32 99b;                   \
        lda $31, $exitin-99b($31);      \
        .previous

#define EXO(x,y...)                     \
        99: x,##y;                      \
        .section __ex_table,"a";        \
        .gprel32 99b;                   \
        lda $31, $exitout-99b($31);     \
        .previous

        .set noat
        .align 4
        .globl __copy_user
        .ent __copy_user
                                # Pipeline info: Slotting & Comments
__copy_user:
        ldgp $29,0($27)         # we do exceptions -- we need the gp.
                                # Macro instruction becomes ldah/lda
                                # .. .. E  E
        .prologue 1
        subq $0, 32, $1         # .. E  .. ..   : Is this going to be a small copy?
        beq $0, $zerolength     # U  .. .. ..   : U L U L

        and $6,7,$3             # .. .. .. E    : is leading dest misalignment
        ble $1, $onebyteloop    # .. .. U  ..   : 1st branch : small amount of data
        beq $3, $destaligned    # .. U  .. ..   : 2nd (one cycle fetcher stall)
        subq $3, 8, $3          # E  .. .. ..   : L U U L : trip counter
/*
 * The fetcher stall also hides the 1 cycle cross-cluster stall for $3 (L --> U)
 * This loop aligns the destination a byte at a time
 * We know we have at least one trip through this loop
 */
$aligndest:
        EXI( ldbu $1,0($7) )    # .. .. .. L    : Keep loads separate from stores
        addq $6,1,$6            # .. .. E  ..   : Section 3.8 in the CWG
        addq $3,1,$3            # .. E  .. ..   :
        nop                     # E  .. .. ..   : U L U L

/*
 * the -1 is to compensate for the inc($6) done in a previous quadpack
 * which allows us zero dependencies within either quadpack in the loop
 */
        EXO( stb $1,-1($6) )    # .. .. .. L    :
        addq $7,1,$7            # .. .. E  ..   : Section 3.8 in the CWG
        subq $0,1,$0            # .. E  .. ..   :
        bne $3, $aligndest      # U  .. .. ..   : U L U L

/*
 * If we fell through into here, we have a minimum of 33 - 7 bytes
 * If we arrived via branch, we have a minimum of 32 bytes
 */
$destaligned:
        and $7,7,$1             # .. .. .. E    : Check _current_ source alignment
        bic $0,7,$4             # .. .. E  ..   : number bytes as a quadword loop
        EXI( ldq_u $3,0($7) )   # .. L  .. ..   : Forward fetch for fallthrough code
        beq $1,$quadaligned     # U  .. .. ..   : U L U L

/*
 * In the worst case, we've just executed an ldq_u here from 0($7)
 * and we'll repeat it once if we take the branch
 */

/* Misaligned quadword loop - not unrolled.  Leave it that way. */
$misquad:
        EXI( ldq_u $2,8($7) )   # .. .. .. L    :
        subq $4,8,$4            # .. .. E  ..   :
        extql $3,$7,$3          # .. U  .. ..   :
        extqh $2,$7,$1          # U  .. .. ..   : U U L L

        bis $3,$1,$1            # .. .. .. E    :
        EXO( stq $1,0($6) )     # .. .. L  ..   :
        addq $7,8,$7            # .. E  .. ..   :
        subq $0,8,$0            # E  .. .. ..   : U L L U

        addq $6,8,$6            # .. .. .. E    :
        bis $2,$2,$3            # .. .. E  ..   :
        nop                     # .. E  .. ..   :
        bne $4,$misquad         # U  .. .. ..   : U L U L

        nop                     # .. .. .. E
        nop                     # .. .. E  ..
        nop                     # .. E  .. ..
        beq $0,$zerolength      # U  .. .. ..   : U L U L

/* We know we have at least one trip through the byte loop */
        EXI ( ldbu $2,0($7) )   # .. .. .. L    : No loads in the same quad
        addq $6,1,$6            # .. .. E  ..   : as the store (Section 3.8 in CWG)
        nop                     # .. E  .. ..   :
        br $31, $dirtyentry     # L0 .. .. ..   : L U U L
/* Do the trailing byte loop load, then hop into the store part of the loop */

/*
 * A minimum of (33 - 7) bytes to do a quad at a time.
 * Based upon the usage context, it's worth the effort to unroll this loop
 * $0 - number of bytes to be moved
 * $4 - number of bytes to move as quadwords
 * $6 is current destination address
 * $7 is current source address
 */
$quadaligned:
        subq    $4, 32, $2      # .. .. .. E    : do not unroll for small stuff
        nop                     # .. .. E  ..
        nop                     # .. E  .. ..
        blt     $2, $onequad    # U  .. .. ..   : U L U L

/*
 * There is a significant assumption here that the source and destination
 * addresses differ by more than 32 bytes.  In this particular case, a
 * sparsity of registers further bounds this to be a minimum of 8 bytes.
 * But if this isn't met, then the output result will be incorrect.
 * Furthermore, due to a lack of available registers, we really can't
 * unroll this to be an 8x loop (which would enable us to use the wh64
 * instruction memory hint instruction).
 */
$unroll4:
        EXI( ldq $1,0($7) )     # .. .. .. L
        EXI( ldq $2,8($7) )     # .. .. L  ..
        subq    $4,32,$4        # .. E  .. ..
        nop                     # E  .. .. ..   : U U L L

        addq    $7,16,$7        # .. .. .. E
        EXO( stq $1,0($6) )     # .. .. L  ..
        EXO( stq $2,8($6) )     # .. L  .. ..
        subq    $0,16,$0        # E  .. .. ..   : U L L U

        addq    $6,16,$6        # .. .. .. E
        EXI( ldq $1,0($7) )     # .. .. L  ..
        EXI( ldq $2,8($7) )     # .. L  .. ..
        subq    $4, 32, $3      # E  .. .. ..   : U U L L : is there enough for another trip?

        EXO( stq $1,0($6) )     # .. .. .. L
        EXO( stq $2,8($6) )     # .. .. L  ..
        subq    $0,16,$0        # .. E  .. ..
        addq    $7,16,$7        # E  .. .. ..   : U L L U

        nop                     # .. .. .. E
        nop                     # .. .. E  ..
        addq    $6,16,$6        # .. E  .. ..
        bgt     $3,$unroll4     # U  .. .. ..   : U L U L

        nop
        nop
        nop
        beq     $4, $noquads

$onequad:
        EXI( ldq $1,0($7) )
        subq    $4,8,$4
        addq    $7,8,$7
        nop

        EXO( stq $1,0($6) )
        subq    $0,8,$0
        addq    $6,8,$6
        bne     $4,$onequad

$noquads:
        nop
        nop
        nop
        beq $0,$zerolength

/*
 * For small copies (or the tail of a larger copy), do a very simple byte loop.
 * There's no point in doing a lot of complex alignment calculations to try to
 * to quadword stuff for a small amount of data.
 *      $0 - remaining number of bytes left to copy
 *      $6 - current dest addr
 *      $7 - current source addr
 */

$onebyteloop:
        EXI ( ldbu $2,0($7) )   # .. .. .. L    : No loads in the same quad
        addq $6,1,$6            # .. .. E  ..   : as the store (Section 3.8 in CWG)
        nop                     # .. E  .. ..   :
        nop                     # E  .. .. ..   : U L U L

$dirtyentry:
/*
 * the -1 is to compensate for the inc($6) done in a previous quadpack
 * which allows us zero dependencies within either quadpack in the loop
 */
        EXO ( stb $2,-1($6) )   # .. .. .. L    :
        addq $7,1,$7            # .. .. E  ..   : quadpack as the load
        subq $0,1,$0            # .. E  .. ..   : change count _after_ copy
        bgt $0,$onebyteloop     # U  .. .. ..   : U L U L

$zerolength:
$exitout:                       # Destination for exception recovery(?)
        nop                     # .. .. .. E
        nop                     # .. .. E  ..
        nop                     # .. E  .. ..
        ret $31,($28),1         # L0 .. .. ..   : L U L U

$exitin:

        /* A stupid byte-by-byte zeroing of the rest of the output
           buffer.  This cures security holes by never leaving 
           random kernel data around to be copied elsewhere.  */

        nop
        nop
        nop
        mov     $0,$1

$101:
        EXO ( stb $31,0($6) )   # L
        subq $1,1,$1            # E
        addq $6,1,$6            # E
        bgt $1,$101             # U

        nop
        nop
        nop
        ret $31,($28),1         # L0

        .end __copy_user