Subversion Repositories or1k_soc_on_altera_embedded_dev_kit

/*

 * arch/alpha/lib/ev6-clear_user.S

 * 21264 version contributed by Rick Gorton 

 *

 * 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";        \

        .long 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:

        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