Adding upstream version 6.1.76.upstream/6.1.76upstream

Signed-off-by: Daniel Baumann <daniel.baumann@progress-linux.org>

1 files changed, 227 insertions, 0 deletions

diff --git a/arch/alpha/lib/ev6-copy_user.S b/arch/alpha/lib/ev6-copy_user.S
new file mode 100644
index 000000000..f3e433754
--- /dev/null
+++ b/arch/alpha/lib/ev6-copy_user.S
@@ -0,0 +1,227 @@
+/* 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 <asm/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)