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authorDaniel Baumann <daniel.baumann@progress-linux.org>2024-04-11 08:27:49 +0000
committerDaniel Baumann <daniel.baumann@progress-linux.org>2024-04-11 08:27:49 +0000
commitace9429bb58fd418f0c81d4c2835699bddf6bde6 (patch)
treeb2d64bc10158fdd5497876388cd68142ca374ed3 /arch/alpha/lib/ev6-copy_user.S
parentInitial commit. (diff)
downloadlinux-ace9429bb58fd418f0c81d4c2835699bddf6bde6.tar.xz
linux-ace9429bb58fd418f0c81d4c2835699bddf6bde6.zip
Adding upstream version 6.6.15.upstream/6.6.15
Signed-off-by: Daniel Baumann <daniel.baumann@progress-linux.org>
Diffstat (limited to 'arch/alpha/lib/ev6-copy_user.S')
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diff --git a/arch/alpha/lib/ev6-copy_user.S b/arch/alpha/lib/ev6-copy_user.S
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+/* SPDX-License-Identifier: GPL-2.0 */
+/*
+ * 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..
+ *
+ * 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
+ */
+
+#include <linux/export.h>
+/* Allow an exception for an insn; exit if we get one. */
+#define EXI(x,y...) \
+ 99: x,##y; \
+ .section __ex_table,"a"; \
+ .long 99b - .; \
+ lda $31, $exitin-99b($31); \
+ .previous
+
+#define EXO(x,y...) \
+ 99: x,##y; \
+ .section __ex_table,"a"; \
+ .long 99b - .; \
+ lda $31, $exitout-99b($31); \
+ .previous
+
+ .set noat
+ .align 4
+ .globl __copy_user
+ .ent __copy_user
+ # Pipeline info: Slotting & Comments
+__copy_user:
+ .prologue 0
+ mov $18, $0 # .. .. .. E
+ subq $18, 32, $1 # .. .. E. .. : Is this going to be a small copy?
+ nop # .. E .. ..
+ beq $18, $zerolength # U .. .. .. : U L U L
+
+ and $16,7,$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($17) ) # .. .. .. L : Keep loads separate from stores
+ addq $16,1,$16 # .. .. 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($16) done in a previous quadpack
+ * which allows us zero dependencies within either quadpack in the loop
+ */
+ EXO( stb $1,-1($16) ) # .. .. .. L :
+ addq $17,1,$17 # .. .. 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 $17,7,$1 # .. .. .. E : Check _current_ source alignment
+ bic $0,7,$4 # .. .. E .. : number bytes as a quadword loop
+ EXI( ldq_u $3,0($17) ) # .. 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($17)
+ * and we'll repeat it once if we take the branch
+ */
+
+/* Misaligned quadword loop - not unrolled. Leave it that way. */
+$misquad:
+ EXI( ldq_u $2,8($17) ) # .. .. .. L :
+ subq $4,8,$4 # .. .. E .. :
+ extql $3,$17,$3 # .. U .. .. :
+ extqh $2,$17,$1 # U .. .. .. : U U L L
+
+ bis $3,$1,$1 # .. .. .. E :
+ EXO( stq $1,0($16) ) # .. .. L .. :
+ addq $17,8,$17 # .. E .. .. :
+ subq $0,8,$0 # E .. .. .. : U L L U
+
+ addq $16,8,$16 # .. .. .. 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($17) ) # .. .. .. L : No loads in the same quad
+ addq $16,1,$16 # .. .. 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
+ * $16 is current destination address
+ * $17 is current source address
+ */
+$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($17) ) # .. .. .. L
+ EXI( ldq $2,8($17) ) # .. .. L ..
+ subq $4,32,$4 # .. E .. ..
+ nop # E .. .. .. : U U L L
+
+ addq $17,16,$17 # .. .. .. E
+ EXO( stq $1,0($16) ) # .. .. L ..
+ EXO( stq $2,8($16) ) # .. L .. ..
+ subq $0,16,$0 # E .. .. .. : U L L U
+
+ addq $16,16,$16 # .. .. .. E
+ EXI( ldq $1,0($17) ) # .. .. L ..
+ EXI( ldq $2,8($17) ) # .. L .. ..
+ subq $4, 32, $3 # E .. .. .. : U U L L : is there enough for another trip?
+
+ EXO( stq $1,0($16) ) # .. .. .. L
+ EXO( stq $2,8($16) ) # .. .. L ..
+ subq $0,16,$0 # .. E .. ..
+ addq $17,16,$17 # E .. .. .. : U L L U
+
+ nop # .. .. .. E
+ nop # .. .. E ..
+ addq $16,16,$16 # .. E .. ..
+ bgt $3,$unroll4 # U .. .. .. : U L U L
+
+ nop
+ nop
+ nop
+ beq $4, $noquads
+
+$onequad:
+ EXI( ldq $1,0($17) )
+ subq $4,8,$4
+ addq $17,8,$17
+ nop
+
+ EXO( stq $1,0($16) )
+ subq $0,8,$0
+ addq $16,8,$16
+ 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
+ * $16 - current dest addr
+ * $17 - current source addr
+ */
+
+$onebyteloop:
+ EXI ( ldbu $2,0($17) ) # .. .. .. L : No loads in the same quad
+ addq $16,1,$16 # .. .. E .. : as the store (Section 3.8 in CWG)
+ nop # .. E .. .. :
+ nop # E .. .. .. : U L U L
+
+$dirtyentry:
+/*
+ * the -1 is to compensate for the inc($16) done in a previous quadpack
+ * which allows us zero dependencies within either quadpack in the loop
+ */
+ EXO ( stb $2,-1($16) ) # .. .. .. L :
+ addq $17,1,$17 # .. .. E .. : quadpack as the load
+ subq $0,1,$0 # .. E .. .. : change count _after_ copy
+ bgt $0,$onebyteloop # U .. .. .. : U L U L
+
+$zerolength:
+$exitin:
+$exitout: # Destination for exception recovery(?)
+ nop # .. .. .. E
+ nop # .. .. E ..
+ nop # .. E .. ..
+ ret $31,($26),1 # L0 .. .. .. : L U L U
+
+ .end __copy_user
+ EXPORT_SYMBOL(__copy_user)