Linux-2.6.12-rc2v2.6.12-rc2

Initial git repository build. I'm not bothering with the full history,
even though we have it. We can create a separate "historical" git
archive of that later if we want to, and in the meantime it's about
3.2GB when imported into git - space that would just make the early
git days unnecessarily complicated, when we don't have a lot of good
infrastructure for it.

Let it rip!

1 files changed, 225 insertions, 0 deletions

diff --git a/arch/alpha/lib/ev6-clear_user.S b/arch/alpha/lib/ev6-clear_user.S
new file mode 100644
index 000000000000..4f42a16b7f53
--- /dev/null
+++ b/arch/alpha/lib/ev6-clear_user.S
@@ -0,0 +1,225 @@
+/*
+ * 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";        \
+        .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
+