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/*
 * arch/alpha/lib/ev6-clear_user.S
 * 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com>
 *
 * Zero user space, handling exceptions as we go.
 *
 * We have to make sure that $0 is always up-to-date and contains the
 * right "bytes left to zero" 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
 *      exception pointer in $7
 *      return address in $28 (exceptions expect it there)
 *
 * Outputs:
 *      bytes left to copy in $0
 *
 * Clobbers:
 *      $1,$2,$3,$4,$5,$6
 *
 * 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
 * Try not to change the actual algorithm if possible for consistency.
 * Determining actual stalls (other than slotting) doesn't appear to be easy to do.
 * From perusing the source code context where this routine is called, it is
 * a fair assumption that significant fractions of entire pages are zeroed, so
 * it's going to be worth the effort to hand-unroll a big loop, and use wh64.
 * ASSUMPTION:
 *      The believed purpose of only updating $0 after a store is that a signal
 *      may come along during the execution of this chunk of code, and we don't
 *      want to leave a hole (and we also want to avoid repeating lots of work)
 */

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

        .set noat
        .set noreorder
        .align 4

        .globl __do_clear_user
        .ent __do_clear_user
        .frame  $30, 0, $28
        .prologue 0

                                # Pipeline info : Slotting & Comments
__do_clear_user:
        ldgp    $29,0($27)      # we do exceptions -- we need the gp.
                                # Macro instruction becomes ldah/lda
                                # .. .. E  E    :
        and     $6, 7, $4       # .. E  .. ..   : find dest head misalignment
        beq     $0, $zerolength # U  .. .. ..   :  U L U L

        addq    $0, $4, $1      # .. .. .. E    : bias counter
        and     $1, 7, $2       # .. .. E  ..   : number of misaligned bytes in tail
# Note - we never actually use $2, so this is a moot computation
# and we can rewrite this later...
        srl     $1, 3, $1       # .. E  .. ..   : number of quadwords to clear
        beq     $4, $headalign  # U  .. .. ..   : U L U L

/*
 * Head is not aligned.  Write (8 - $4) bytes to head of destination
 * This means $6 is known to be misaligned
 */
        EX( ldq_u $5, 0($6) )   # .. .. .. L    : load dst word to mask back in
        beq     $1, $onebyte    # .. .. U  ..   : sub-word store?
        mskql   $5, $6, $5      # .. U  .. ..   : take care of misaligned head
        addq    $6, 8, $6       # E  .. .. ..   : L U U L

        EX( stq_u $5, -8($6) )  # .. .. .. L    :
        subq    $1, 1, $1       # .. .. E  ..   :
        addq    $0, $4, $0      # .. E  .. ..   : bytes left -= 8 - misalignment
        subq    $0, 8, $0       # E  .. .. ..   : U L U L

        .align  4
/*
 * (The .align directive ought to be a moot point)
 * values upon initial entry to the loop
 * $1 is number of quadwords to clear (zero is a valid value)
 * $2 is number of trailing bytes (0..7) ($2 never used...)
 * $6 is known to be aligned 0mod8
 */
$headalign:
        subq    $1, 16, $4      # .. .. .. E    : If < 16, we can not use the huge loop
        and     $6, 0x3f, $2    # .. .. E  ..   : Forward work for huge loop
        subq    $2, 0x40, $3    # .. E  .. ..   : bias counter (huge loop)
        blt     $4, $trailquad  # U  .. .. ..   : U L U L

/*
 * We know that we're going to do at least 16 quads, which means we are
 * going to be able to use the large block clear loop at least once.
 * Figure out how many quads we need to clear before we are 0mod64 aligned
 * so we can use the wh64 instruction.
 */

        nop                     # .. .. .. E
        nop                     # .. .. E  ..
        nop                     # .. E  .. ..
        beq     $3, $bigalign   # U  .. .. ..   : U L U L : Aligned 0mod64

$alignmod64:
        EX( stq_u $31, 0($6) )  # .. .. .. L
        addq    $3, 8, $3       # .. .. E  ..
        subq    $0, 8, $0       # .. E  .. ..
        nop                     # E  .. .. ..   : U L U L

        nop                     # .. .. .. E
        subq    $1, 1, $1       # .. .. E  ..
        addq    $6, 8, $6       # .. E  .. ..
        blt     $3, $alignmod64 # U  .. .. ..   : U L U L

$bigalign:
/*
 * $0 is the number of bytes left
 * $1 is the number of quads left
 * $6 is aligned 0mod64
 * we know that we'll be taking a minimum of one trip through
 * CWG Section 3.7.6: do not expect a sustained store rate of > 1/cycle
 * We are _not_ going to update $0 after every single store.  That
 * would be silly, because there will be cross-cluster dependencies
 * no matter how the code is scheduled.  By doing it in slightly
 * staggered fashion, we can still do this loop in 5 fetches
 * The worse case will be doing two extra quads in some future execution,
 * in the event of an interrupted clear.
 * Assumes the wh64 needs to be for 2 trips through the loop in the future
 * The wh64 is issued on for the starting destination address for trip +2
 * through the loop, and if there are less than two trips left, the target
 * address will be for the current trip.
 */
        nop                     # E :
        nop                     # E :
        nop                     # E :
        bis     $6,$6,$3        # E : U L U L : Initial wh64 address is dest
        /* This might actually help for the current trip... */

$do_wh64:
        wh64    ($3)            # .. .. .. L1   : memory subsystem hint
        subq    $1, 16, $4      # .. .. E  ..   : Forward calculation - repeat the loop?
        EX( stq_u $31, 0($6) )  # .. L  .. ..
        subq    $0, 8, $0       # E  .. .. ..   : U L U L

        addq    $6, 128, $3     # E : Target address of wh64
        EX( stq_u $31, 8($6) )  # L :
        EX( stq_u $31, 16($6) ) # L :
        subq    $0, 16, $0      # E : U L L U

        nop                     # E :
        EX( stq_u $31, 24($6) ) # L :
        EX( stq_u $31, 32($6) ) # L :
        subq    $0, 168, $5     # E : U L L U : two trips through the loop left?
        /* 168 = 192 - 24, since we've already completed some stores */

        subq    $0, 16, $0      # E :
        EX( stq_u $31, 40($6) ) # L :
        EX( stq_u $31, 48($6) ) # L :
        cmovlt  $5, $6, $3      # E : U L L U : Latency 2, extra mapping cycle

        subq    $1, 8, $1       # E :
        subq    $0, 16, $0      # E :
        EX( stq_u $31, 56($6) ) # L :
        nop                     # E : U L U L

        nop                     # E :
        subq    $0, 8, $0       # E :
        addq    $6, 64, $6      # E :
        bge     $4, $do_wh64    # U : U L U L

$trailquad:
        # zero to 16 quadwords left to store, plus any trailing bytes
        # $1 is the number of quadwords left to go.
        # 
        nop                     # .. .. .. E
        nop                     # .. .. E  ..
        nop                     # .. E  .. ..
        beq     $1, $trailbytes # U  .. .. ..   : U L U L : Only 0..7 bytes to go

$onequad:
        EX( stq_u $31, 0($6) )  # .. .. .. L
        subq    $1, 1, $1       # .. .. E  ..
        subq    $0, 8, $0       # .. E  .. ..
        nop                     # E  .. .. ..   : U L U L

        nop                     # .. .. .. E
        nop                     # .. .. E  ..
        addq    $6, 8, $6       # .. E  .. ..
        bgt     $1, $onequad    # U  .. .. ..   : U L U L

        # We have an unknown number of bytes left to go.
$trailbytes:
        nop                     # .. .. .. E
        nop                     # .. .. E  ..
        nop                     # .. E  .. ..
        beq     $0, $zerolength # U  .. .. ..   : U L U L

        # $0 contains the number of bytes left to copy (0..31)
        # so we will use $0 as the loop counter
        # We know for a fact that $0 > 0 zero due to previous context
$onebyte:
        EX( stb $31, 0($6) )    # .. .. .. L
        subq    $0, 1, $0       # .. .. E  ..   :
        addq    $6, 1, $6       # .. E  .. ..   :
        bgt     $0, $onebyte    # U  .. .. ..   : U L U L

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