192 lines
5.2 KiB
ArmAsm
192 lines
5.2 KiB
ArmAsm
/*
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* arch/alpha/lib/ev6-memchr.S
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*
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* 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com>
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*
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* Finds characters in a memory area. Optimized for the Alpha:
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*
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* - memory accessed as aligned quadwords only
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* - uses cmpbge to compare 8 bytes in parallel
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* - does binary search to find 0 byte in last
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* quadword (HAKMEM needed 12 instructions to
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* do this instead of the 9 instructions that
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* binary search needs).
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*
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* For correctness consider that:
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*
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* - only minimum number of quadwords may be accessed
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* - the third argument is an unsigned long
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*
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* Much of the information about 21264 scheduling/coding comes from:
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* Compiler Writer's Guide for the Alpha 21264
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* abbreviated as 'CWG' in other comments here
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* ftp.digital.com/pub/Digital/info/semiconductor/literature/dsc-library.html
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* Scheduling notation:
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* E - either cluster
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* U - upper subcluster; U0 - subcluster U0; U1 - subcluster U1
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* L - lower subcluster; L0 - subcluster L0; L1 - subcluster L1
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* Try not to change the actual algorithm if possible for consistency.
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*/
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.set noreorder
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.set noat
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.align 4
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.globl memchr
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.ent memchr
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memchr:
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.frame $30,0,$26,0
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.prologue 0
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# Hack -- if someone passes in (size_t)-1, hoping to just
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# search til the end of the address space, we will overflow
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# below when we find the address of the last byte. Given
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# that we will never have a 56-bit address space, cropping
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# the length is the easiest way to avoid trouble.
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zap $18, 0x80, $5 # U : Bound length
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beq $18, $not_found # U :
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ldq_u $1, 0($16) # L : load first quadword Latency=3
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and $17, 0xff, $17 # E : L L U U : 00000000000000ch
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insbl $17, 1, $2 # U : 000000000000ch00
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cmpult $18, 9, $4 # E : small (< 1 quad) string?
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or $2, $17, $17 # E : 000000000000chch
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lda $3, -1($31) # E : U L L U
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sll $17, 16, $2 # U : 00000000chch0000
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addq $16, $5, $5 # E : Max search address
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or $2, $17, $17 # E : 00000000chchchch
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sll $17, 32, $2 # U : U L L U : chchchch00000000
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or $2, $17, $17 # E : chchchchchchchch
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extql $1, $16, $7 # U : $7 is upper bits
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beq $4, $first_quad # U :
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ldq_u $6, -1($5) # L : L U U L : eight or less bytes to search Latency=3
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extqh $6, $16, $6 # U : 2 cycle stall for $6
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mov $16, $0 # E :
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nop # E :
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or $7, $6, $1 # E : L U L U $1 = quadword starting at $16
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# Deal with the case where at most 8 bytes remain to be searched
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# in $1. E.g.:
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# $18 = 6
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# $1 = ????c6c5c4c3c2c1
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$last_quad:
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negq $18, $6 # E :
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xor $17, $1, $1 # E :
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srl $3, $6, $6 # U : $6 = mask of $18 bits set
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cmpbge $31, $1, $2 # E : L U L U
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nop
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nop
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and $2, $6, $2 # E :
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beq $2, $not_found # U : U L U L
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$found_it:
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#ifdef CONFIG_ALPHA_EV67
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/*
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* Since we are guaranteed to have set one of the bits, we don't
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* have to worry about coming back with a 0x40 out of cttz...
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*/
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cttz $2, $3 # U0 :
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addq $0, $3, $0 # E : All done
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nop # E :
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ret # L0 : L U L U
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#else
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/*
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* Slow and clunky. It can probably be improved.
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* An exercise left for others.
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*/
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negq $2, $3 # E :
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and $2, $3, $2 # E :
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and $2, 0x0f, $1 # E :
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addq $0, 4, $3 # E :
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cmoveq $1, $3, $0 # E : Latency 2, extra map cycle
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nop # E : keep with cmov
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and $2, 0x33, $1 # E :
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addq $0, 2, $3 # E : U L U L : 2 cycle stall on $0
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cmoveq $1, $3, $0 # E : Latency 2, extra map cycle
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nop # E : keep with cmov
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and $2, 0x55, $1 # E :
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addq $0, 1, $3 # E : U L U L : 2 cycle stall on $0
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cmoveq $1, $3, $0 # E : Latency 2, extra map cycle
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nop
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nop
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ret # L0 : L U L U
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#endif
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# Deal with the case where $18 > 8 bytes remain to be
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# searched. $16 may not be aligned.
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.align 4
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$first_quad:
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andnot $16, 0x7, $0 # E :
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insqh $3, $16, $2 # U : $2 = 0000ffffffffffff ($16<0:2> ff)
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xor $1, $17, $1 # E :
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or $1, $2, $1 # E : U L U L $1 = ====ffffffffffff
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cmpbge $31, $1, $2 # E :
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bne $2, $found_it # U :
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# At least one byte left to process.
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ldq $1, 8($0) # L :
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subq $5, 1, $18 # E : U L U L
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addq $0, 8, $0 # E :
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# Make $18 point to last quad to be accessed (the
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# last quad may or may not be partial).
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andnot $18, 0x7, $18 # E :
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cmpult $0, $18, $2 # E :
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beq $2, $final # U : U L U L
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# At least two quads remain to be accessed.
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subq $18, $0, $4 # E : $4 <- nr quads to be processed
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and $4, 8, $4 # E : odd number of quads?
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bne $4, $odd_quad_count # U :
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# At least three quads remain to be accessed
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mov $1, $4 # E : L U L U : move prefetched value to correct reg
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.align 4
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$unrolled_loop:
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ldq $1, 8($0) # L : prefetch $1
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xor $17, $4, $2 # E :
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cmpbge $31, $2, $2 # E :
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bne $2, $found_it # U : U L U L
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addq $0, 8, $0 # E :
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nop # E :
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nop # E :
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nop # E :
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$odd_quad_count:
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xor $17, $1, $2 # E :
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ldq $4, 8($0) # L : prefetch $4
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cmpbge $31, $2, $2 # E :
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addq $0, 8, $6 # E :
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bne $2, $found_it # U :
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cmpult $6, $18, $6 # E :
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addq $0, 8, $0 # E :
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nop # E :
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bne $6, $unrolled_loop # U :
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mov $4, $1 # E : move prefetched value into $1
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nop # E :
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nop # E :
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$final: subq $5, $0, $18 # E : $18 <- number of bytes left to do
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nop # E :
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nop # E :
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bne $18, $last_quad # U :
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$not_found:
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mov $31, $0 # E :
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nop # E :
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nop # E :
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ret # L0 :
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.end memchr
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