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kernel/linux-imx6_3.14.28/arch/alpha/lib/ev6-memchr.S 5.23 KB
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  /*
   * arch/alpha/lib/ev6-memchr.S
   *
   * 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com>
   *
   * Finds characters in a memory area.  Optimized for the Alpha:
   *
   *    - memory accessed as aligned quadwords only
   *    - uses cmpbge to compare 8 bytes in parallel
   *    - does binary search to find 0 byte in last
   *      quadword (HAKMEM needed 12 instructions to
   *      do this instead of the 9 instructions that
   *      binary search needs).
   *
   * For correctness consider that:
   *
   *    - only minimum number of quadwords may be accessed
   *    - the third argument is an unsigned long
   *
   * 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.
   */
  
          .set noreorder
          .set noat
  
  	.align	4
  	.globl memchr
  	.ent memchr
  memchr:
  	.frame $30,0,$26,0
  	.prologue 0
  
  	# Hack -- if someone passes in (size_t)-1, hoping to just
  	# search til the end of the address space, we will overflow
  	# below when we find the address of the last byte.  Given
  	# that we will never have a 56-bit address space, cropping
  	# the length is the easiest way to avoid trouble.
  	zap	$18, 0x80, $5	# U : Bound length
  	beq	$18, $not_found	# U :
          ldq_u   $1, 0($16)	# L : load first quadword Latency=3
  	and	$17, 0xff, $17	# E : L L U U : 00000000000000ch
  
  	insbl	$17, 1, $2	# U : 000000000000ch00
  	cmpult	$18, 9, $4	# E : small (< 1 quad) string?
  	or	$2, $17, $17	# E : 000000000000chch
          lda     $3, -1($31)	# E : U L L U
  
  	sll	$17, 16, $2	# U : 00000000chch0000
  	addq	$16, $5, $5	# E : Max search address
  	or	$2, $17, $17	# E : 00000000chchchch
  	sll	$17, 32, $2	# U : U L L U : chchchch00000000
  
  	or	$2, $17, $17	# E : chchchchchchchch
  	extql	$1, $16, $7	# U : $7 is upper bits
  	beq	$4, $first_quad	# U :
  	ldq_u	$6, -1($5)	# L : L U U L : eight or less bytes to search Latency=3
  
  	extqh	$6, $16, $6	# U : 2 cycle stall for $6
  	mov	$16, $0		# E :
  	nop			# E :
  	or	$7, $6, $1	# E : L U L U $1 = quadword starting at $16
  
  	# Deal with the case where at most 8 bytes remain to be searched
  	# in $1.  E.g.:
  	#	$18 = 6
  	#	$1 = ????c6c5c4c3c2c1
  $last_quad:
  	negq	$18, $6		# E :
          xor	$17, $1, $1	# E :
  	srl	$3, $6, $6	# U : $6 = mask of $18 bits set
          cmpbge  $31, $1, $2	# E : L U L U
  
  	nop
  	nop
  	and	$2, $6, $2	# E :
          beq     $2, $not_found	# U : U L U L
  
  $found_it:
  #ifdef CONFIG_ALPHA_EV67
  	/*
  	 * Since we are guaranteed to have set one of the bits, we don't
  	 * have to worry about coming back with a 0x40 out of cttz...
  	 */
  	cttz	$2, $3		# U0 :
  	addq	$0, $3, $0	# E : All done
  	nop			# E :
  	ret			# L0 : L U L U
  #else
  	/*
  	 * Slow and clunky.  It can probably be improved.
  	 * An exercise left for others.
  	 */
          negq    $2, $3		# E :
          and     $2, $3, $2	# E :
          and     $2, 0x0f, $1	# E :
          addq    $0, 4, $3	# E :
  
          cmoveq  $1, $3, $0	# E : Latency 2, extra map cycle
  	nop			# E : keep with cmov
          and     $2, 0x33, $1	# E :
          addq    $0, 2, $3	# E : U L U L : 2 cycle stall on $0
  
          cmoveq  $1, $3, $0	# E : Latency 2, extra map cycle
  	nop			# E : keep with cmov
          and     $2, 0x55, $1	# E :
          addq    $0, 1, $3	# E : U L U L : 2 cycle stall on $0
  
          cmoveq  $1, $3, $0	# E : Latency 2, extra map cycle
  	nop
  	nop
  	ret			# L0 : L U L U
  #endif
  
  	# Deal with the case where $18 > 8 bytes remain to be
  	# searched.  $16 may not be aligned.
  	.align 4
  $first_quad:
  	andnot	$16, 0x7, $0	# E :
          insqh   $3, $16, $2	# U : $2 = 0000ffffffffffff ($16<0:2> ff)
          xor	$1, $17, $1	# E :
  	or	$1, $2, $1	# E : U L U L $1 = ====ffffffffffff
  
          cmpbge  $31, $1, $2	# E :
          bne     $2, $found_it	# U :
  	# At least one byte left to process.
  	ldq	$1, 8($0)	# L :
  	subq	$5, 1, $18	# E : U L U L
  
  	addq	$0, 8, $0	# E :
  	# Make $18 point to last quad to be accessed (the
  	# last quad may or may not be partial).
  	andnot	$18, 0x7, $18	# E :
  	cmpult	$0, $18, $2	# E :
  	beq	$2, $final	# U : U L U L
  
  	# At least two quads remain to be accessed.
  
  	subq	$18, $0, $4	# E : $4 <- nr quads to be processed
  	and	$4, 8, $4	# E : odd number of quads?
  	bne	$4, $odd_quad_count # U :
  	# At least three quads remain to be accessed
  	mov	$1, $4		# E : L U L U : move prefetched value to correct reg
  
  	.align	4
  $unrolled_loop:
  	ldq	$1, 8($0)	# L : prefetch $1
  	xor	$17, $4, $2	# E :
  	cmpbge	$31, $2, $2	# E :
  	bne	$2, $found_it	# U : U L U L
  
  	addq	$0, 8, $0	# E :
  	nop			# E :
  	nop			# E :
  	nop			# E :
  
  $odd_quad_count:
  	xor	$17, $1, $2	# E :
  	ldq	$4, 8($0)	# L : prefetch $4
  	cmpbge	$31, $2, $2	# E :
  	addq	$0, 8, $6	# E :
  
  	bne	$2, $found_it	# U :
  	cmpult	$6, $18, $6	# E :
  	addq	$0, 8, $0	# E :
  	nop			# E :
  
  	bne	$6, $unrolled_loop # U :
  	mov	$4, $1		# E : move prefetched value into $1
  	nop			# E :
  	nop			# E :
  
  $final:	subq	$5, $0, $18	# E : $18 <- number of bytes left to do
  	nop			# E :
  	nop			# E :
  	bne	$18, $last_quad	# U :
  
  $not_found:
  	mov	$31, $0		# E :
  	nop			# E :
  	nop			# E :
  	ret			# L0 :
  
          .end memchr