alpha/lib/ev6-clear_user.S

8  * We have to make sure that $0 is always up-to-date and contains the
18  *	E	- either cluster
27  *	The believed purpose of only updating $0 after a store is that a signal
47 	.frame	$30, 0, $26
48 	.prologue 0
52 	and	$17, $17, $0
53 	and	$16, 7, $4	# .. E  .. ..	: find dest head misalignment
54 	beq	$0, $zerolength # U  .. .. ..	:  U L U L
56 	addq	$0, $4, $1	# .. .. .. E	: bias counter
57 	and	$1, 7, $2	# .. .. E  ..	: number of misaligned bytes in tail
60 	srl	$1, 3, $1	# .. E  .. ..	: number of quadwords to clear
67 	EX( ldq_u $5, 0($16) )	# .. .. .. L	: load dst word to mask back in
70 	addq	$16, 8, $16	# E  .. .. .. 	: L U U L
73 	subq	$1, 1, $1	# .. .. E  ..	:
74 	addq	$0, $4, $0	# .. E  .. ..	: bytes left -= 8 - misalignment
75 	subq	$0, 8, $0	# E  .. .. ..	: U L U L
82  * $2 is number of trailing bytes (0..7) ($2 never used...)
83  * $16 is known to be aligned 0mod8
86 	subq	$1, 16, $4	# .. .. .. E	: If < 16, we can not use the huge loop
87 	and	$16, 0x3f, $2	# .. .. E  ..	: Forward work for huge loop
88 	subq	$2, 0x40, $3	# .. E  .. ..	: bias counter (huge loop)
94  * Figure out how many quads we need to clear before we are 0mod64 aligned
98 	nop			# .. .. .. E
99 	nop			# .. .. E  ..
100 	nop			# .. E  .. ..
101 	beq	$3, $bigalign	# U  .. .. ..	: U L U L : Aligned 0mod64
104 	EX( stq_u $31, 0($16) )	# .. .. .. L
105 	addq	$3, 8, $3	# .. .. E  ..
106 	subq	$0, 8, $0	# .. E  .. ..
107 	nop			# E  .. .. ..	: U L U L
109 	nop			# .. .. .. E
110 	subq	$1, 1, $1	# .. .. E  ..
111 	addq	$16, 8, $16	# .. E  .. ..
116  * $0 is the number of bytes left
118  * $16 is aligned 0mod64
121  * We are _not_ going to update $0 after every single store.  That
132 	nop			# E :
133 	nop			# E :
134 	nop			# E :
135 	bis	$16,$16,$3	# E : U L U L : Initial wh64 address is dest
140 	subq	$1, 16, $4	# .. .. E  ..	: Forward calculation - repeat the loop?
141 	EX( stq_u $31, 0($16) )	# .. L  .. ..
142 	subq	$0, 8, $0	# E  .. .. ..	: U L U L
144 	addq	$16, 128, $3	# E : Target address of wh64
147 	subq	$0, 16, $0	# E : U L L U
149 	nop			# E :
152 	subq	$0, 168, $5	# E : U L L U : two trips through the loop left?
155 	subq	$0, 16, $0	# E :
158 	cmovlt	$5, $16, $3	# E : U L L U : Latency 2, extra mapping cycle
160 	subq	$1, 8, $1	# E :
161 	subq	$0, 16, $0	# E :
163 	nop			# E : U L U L
165 	nop			# E :
166 	subq	$0, 8, $0	# E :
167 	addq	$16, 64, $16	# E :
174 	nop			# .. .. .. E
175 	nop			# .. .. E  ..
176 	nop			# .. E  .. ..
177 	beq	$1, $trailbytes	# U  .. .. ..	: U L U L : Only 0..7 bytes to go
180 	EX( stq_u $31, 0($16) )	# .. .. .. L
181 	subq	$1, 1, $1	# .. .. E  ..
182 	subq	$0, 8, $0	# .. E  .. ..
183 	nop			# E  .. .. ..	: U L U L
185 	nop			# .. .. .. E
186 	nop			# .. .. E  ..
187 	addq	$16, 8, $16	# .. E  .. ..
192 	nop			# .. .. .. E
193 	nop			# .. .. E  ..
194 	nop			# .. E  .. ..
195 	beq	$0, $zerolength	# U  .. .. ..	: U L U L
197 	# $0 contains the number of bytes left to copy (0..31)
198 	# so we will use $0 as the loop counter
199 	# We know for a fact that $0 > 0 zero due to previous context
201 	EX( stb $31, 0($16) )	# .. .. .. L
202 	subq	$0, 1, $0	# .. .. E  ..	:
203 	addq	$16, 1, $16	# .. E  .. ..	:
204 	bgt	$0, $onebyte	# U  .. .. ..	: U L U L
208 	nop			# .. .. .. E	:
209 	nop			# .. .. E  ..	:
210 	nop			# .. E  .. ..	: