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authorAndré Fabian Silva Delgado <emulatorman@parabola.nu>2015-08-05 17:04:01 -0300
committerAndré Fabian Silva Delgado <emulatorman@parabola.nu>2015-08-05 17:04:01 -0300
commit57f0f512b273f60d52568b8c6b77e17f5636edc0 (patch)
tree5e910f0e82173f4ef4f51111366a3f1299037a7b /arch/alpha/lib/ev6-clear_user.S
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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"; \
+ .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
+