605 lines
16 KiB
ArmAsm
605 lines
16 KiB
ArmAsm
/*
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* arch/alpha/lib/ev6-memset.S
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*
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* This is an efficient (and relatively small) implementation of the C library
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* "memset()" function for the 21264 implementation of Alpha.
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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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* 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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* The algorithm for the leading and trailing quadwords remains the same,
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* however the loop has been unrolled to enable better memory throughput,
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* and the code has been replicated for each of the entry points: __memset
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* and __memsetw to permit better scheduling to eliminate the stalling
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* encountered during the mask replication.
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* A future enhancement might be to put in a byte store loop for really
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* small (say < 32 bytes) memset()s. Whether or not that change would be
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* a win in the kernel would depend upon the contextual usage.
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* WARNING: Maintaining this is going to be more work than the above version,
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* as fixes will need to be made in multiple places. The performance gain
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* is worth it.
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*/
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#include <asm/export.h>
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.set noat
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.set noreorder
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.text
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.globl memset
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.globl __memset
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.globl ___memset
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.globl __memsetw
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.globl __constant_c_memset
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.ent ___memset
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.align 5
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___memset:
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.frame $30,0,$26,0
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.prologue 0
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/*
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* Serious stalling happens. The only way to mitigate this is to
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* undertake a major re-write to interleave the constant materialization
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* with other parts of the fall-through code. This is important, even
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* though it makes maintenance tougher.
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* Do this later.
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*/
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and $17,255,$1 # E : 00000000000000ch
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insbl $17,1,$2 # U : 000000000000ch00
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bis $16,$16,$0 # E : return value
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ble $18,end_b # U : zero length requested?
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addq $18,$16,$6 # E : max address to write to
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bis $1,$2,$17 # E : 000000000000chch
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insbl $1,2,$3 # U : 0000000000ch0000
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insbl $1,3,$4 # U : 00000000ch000000
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or $3,$4,$3 # E : 00000000chch0000
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inswl $17,4,$5 # U : 0000chch00000000
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xor $16,$6,$1 # E : will complete write be within one quadword?
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inswl $17,6,$2 # U : chch000000000000
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or $17,$3,$17 # E : 00000000chchchch
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or $2,$5,$2 # E : chchchch00000000
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bic $1,7,$1 # E : fit within a single quadword?
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and $16,7,$3 # E : Target addr misalignment
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or $17,$2,$17 # E : chchchchchchchch
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beq $1,within_quad_b # U :
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nop # E :
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beq $3,aligned_b # U : target is 0mod8
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/*
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* Target address is misaligned, and won't fit within a quadword
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*/
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ldq_u $4,0($16) # L : Fetch first partial
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bis $16,$16,$5 # E : Save the address
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insql $17,$16,$2 # U : Insert new bytes
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subq $3,8,$3 # E : Invert (for addressing uses)
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addq $18,$3,$18 # E : $18 is new count ($3 is negative)
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mskql $4,$16,$4 # U : clear relevant parts of the quad
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subq $16,$3,$16 # E : $16 is new aligned destination
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bis $2,$4,$1 # E : Final bytes
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nop
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stq_u $1,0($5) # L : Store result
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nop
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nop
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.align 4
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aligned_b:
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/*
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* We are now guaranteed to be quad aligned, with at least
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* one partial quad to write.
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*/
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sra $18,3,$3 # U : Number of remaining quads to write
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and $18,7,$18 # E : Number of trailing bytes to write
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bis $16,$16,$5 # E : Save dest address
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beq $3,no_quad_b # U : tail stuff only
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/*
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* it's worth the effort to unroll this and use wh64 if possible
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* Lifted a bunch of code from clear_user.S
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* At this point, entry values are:
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* $16 Current destination address
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* $5 A copy of $16
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* $6 The max quadword address to write to
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* $18 Number trailer bytes
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* $3 Number quads to write
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*/
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and $16, 0x3f, $2 # E : Forward work (only useful for unrolled loop)
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subq $3, 16, $4 # E : Only try to unroll if > 128 bytes
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subq $2, 0x40, $1 # E : bias counter (aligning stuff 0mod64)
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blt $4, loop_b # U :
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/*
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* We know we've got at least 16 quads, minimum of one trip
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* through unrolled loop. Do a quad at a time to get us 0mod64
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* aligned.
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*/
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nop # E :
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nop # E :
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nop # E :
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beq $1, $bigalign_b # U :
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$alignmod64_b:
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stq $17, 0($5) # L :
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subq $3, 1, $3 # E : For consistency later
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addq $1, 8, $1 # E : Increment towards zero for alignment
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addq $5, 8, $4 # E : Initial wh64 address (filler instruction)
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nop
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nop
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addq $5, 8, $5 # E : Inc address
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blt $1, $alignmod64_b # U :
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$bigalign_b:
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/*
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* $3 - number quads left to go
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* $5 - target address (aligned 0mod64)
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* $17 - mask of stuff to store
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* Scratch registers available: $7, $2, $4, $1
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* we know that we'll be taking a minimum of one trip through
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* CWG Section 3.7.6: do not expect a sustained store rate of > 1/cycle
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* Assumes the wh64 needs to be for 2 trips through the loop in the future
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* The wh64 is issued on for the starting destination address for trip +2
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* through the loop, and if there are less than two trips left, the target
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* address will be for the current trip.
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*/
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$do_wh64_b:
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wh64 ($4) # L1 : memory subsystem write hint
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subq $3, 24, $2 # E : For determining future wh64 addresses
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stq $17, 0($5) # L :
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nop # E :
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addq $5, 128, $4 # E : speculative target of next wh64
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stq $17, 8($5) # L :
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stq $17, 16($5) # L :
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addq $5, 64, $7 # E : Fallback address for wh64 (== next trip addr)
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stq $17, 24($5) # L :
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stq $17, 32($5) # L :
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cmovlt $2, $7, $4 # E : Latency 2, extra mapping cycle
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nop
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stq $17, 40($5) # L :
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stq $17, 48($5) # L :
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subq $3, 16, $2 # E : Repeat the loop at least once more?
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nop
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stq $17, 56($5) # L :
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addq $5, 64, $5 # E :
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subq $3, 8, $3 # E :
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bge $2, $do_wh64_b # U :
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nop
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nop
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nop
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beq $3, no_quad_b # U : Might have finished already
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.align 4
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/*
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* Simple loop for trailing quadwords, or for small amounts
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* of data (where we can't use an unrolled loop and wh64)
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*/
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loop_b:
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stq $17,0($5) # L :
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subq $3,1,$3 # E : Decrement number quads left
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addq $5,8,$5 # E : Inc address
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bne $3,loop_b # U : more?
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no_quad_b:
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/*
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* Write 0..7 trailing bytes.
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*/
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nop # E :
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beq $18,end_b # U : All done?
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ldq $7,0($5) # L :
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mskqh $7,$6,$2 # U : Mask final quad
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insqh $17,$6,$4 # U : New bits
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bis $2,$4,$1 # E : Put it all together
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stq $1,0($5) # L : And back to memory
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ret $31,($26),1 # L0 :
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within_quad_b:
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ldq_u $1,0($16) # L :
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insql $17,$16,$2 # U : New bits
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mskql $1,$16,$4 # U : Clear old
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bis $2,$4,$2 # E : New result
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mskql $2,$6,$4 # U :
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mskqh $1,$6,$2 # U :
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bis $2,$4,$1 # E :
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stq_u $1,0($16) # L :
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end_b:
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nop
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nop
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nop
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ret $31,($26),1 # L0 :
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.end ___memset
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EXPORT_SYMBOL(___memset)
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/*
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* This is the original body of code, prior to replication and
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* rescheduling. Leave it here, as there may be calls to this
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* entry point.
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*/
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.align 4
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.ent __constant_c_memset
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__constant_c_memset:
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.frame $30,0,$26,0
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.prologue 0
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addq $18,$16,$6 # E : max address to write to
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bis $16,$16,$0 # E : return value
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xor $16,$6,$1 # E : will complete write be within one quadword?
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ble $18,end # U : zero length requested?
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bic $1,7,$1 # E : fit within a single quadword
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beq $1,within_one_quad # U :
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and $16,7,$3 # E : Target addr misalignment
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beq $3,aligned # U : target is 0mod8
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/*
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* Target address is misaligned, and won't fit within a quadword
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*/
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ldq_u $4,0($16) # L : Fetch first partial
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bis $16,$16,$5 # E : Save the address
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insql $17,$16,$2 # U : Insert new bytes
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subq $3,8,$3 # E : Invert (for addressing uses)
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addq $18,$3,$18 # E : $18 is new count ($3 is negative)
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mskql $4,$16,$4 # U : clear relevant parts of the quad
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subq $16,$3,$16 # E : $16 is new aligned destination
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bis $2,$4,$1 # E : Final bytes
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nop
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stq_u $1,0($5) # L : Store result
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nop
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nop
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.align 4
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aligned:
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/*
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* We are now guaranteed to be quad aligned, with at least
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* one partial quad to write.
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*/
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sra $18,3,$3 # U : Number of remaining quads to write
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and $18,7,$18 # E : Number of trailing bytes to write
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bis $16,$16,$5 # E : Save dest address
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beq $3,no_quad # U : tail stuff only
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/*
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* it's worth the effort to unroll this and use wh64 if possible
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* Lifted a bunch of code from clear_user.S
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* At this point, entry values are:
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* $16 Current destination address
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* $5 A copy of $16
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* $6 The max quadword address to write to
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* $18 Number trailer bytes
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* $3 Number quads to write
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*/
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and $16, 0x3f, $2 # E : Forward work (only useful for unrolled loop)
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subq $3, 16, $4 # E : Only try to unroll if > 128 bytes
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subq $2, 0x40, $1 # E : bias counter (aligning stuff 0mod64)
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blt $4, loop # U :
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/*
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* We know we've got at least 16 quads, minimum of one trip
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* through unrolled loop. Do a quad at a time to get us 0mod64
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* aligned.
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*/
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nop # E :
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nop # E :
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nop # E :
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beq $1, $bigalign # U :
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$alignmod64:
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stq $17, 0($5) # L :
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subq $3, 1, $3 # E : For consistency later
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addq $1, 8, $1 # E : Increment towards zero for alignment
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addq $5, 8, $4 # E : Initial wh64 address (filler instruction)
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nop
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nop
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addq $5, 8, $5 # E : Inc address
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blt $1, $alignmod64 # U :
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$bigalign:
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/*
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* $3 - number quads left to go
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* $5 - target address (aligned 0mod64)
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* $17 - mask of stuff to store
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* Scratch registers available: $7, $2, $4, $1
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* we know that we'll be taking a minimum of one trip through
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* CWG Section 3.7.6: do not expect a sustained store rate of > 1/cycle
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* Assumes the wh64 needs to be for 2 trips through the loop in the future
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* The wh64 is issued on for the starting destination address for trip +2
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* through the loop, and if there are less than two trips left, the target
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* address will be for the current trip.
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*/
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$do_wh64:
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wh64 ($4) # L1 : memory subsystem write hint
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subq $3, 24, $2 # E : For determining future wh64 addresses
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stq $17, 0($5) # L :
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nop # E :
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addq $5, 128, $4 # E : speculative target of next wh64
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stq $17, 8($5) # L :
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stq $17, 16($5) # L :
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addq $5, 64, $7 # E : Fallback address for wh64 (== next trip addr)
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stq $17, 24($5) # L :
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stq $17, 32($5) # L :
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cmovlt $2, $7, $4 # E : Latency 2, extra mapping cycle
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nop
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stq $17, 40($5) # L :
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stq $17, 48($5) # L :
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subq $3, 16, $2 # E : Repeat the loop at least once more?
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nop
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stq $17, 56($5) # L :
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addq $5, 64, $5 # E :
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subq $3, 8, $3 # E :
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bge $2, $do_wh64 # U :
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nop
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nop
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nop
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beq $3, no_quad # U : Might have finished already
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.align 4
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/*
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* Simple loop for trailing quadwords, or for small amounts
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* of data (where we can't use an unrolled loop and wh64)
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*/
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loop:
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stq $17,0($5) # L :
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subq $3,1,$3 # E : Decrement number quads left
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addq $5,8,$5 # E : Inc address
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bne $3,loop # U : more?
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no_quad:
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/*
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* Write 0..7 trailing bytes.
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*/
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nop # E :
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beq $18,end # U : All done?
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ldq $7,0($5) # L :
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mskqh $7,$6,$2 # U : Mask final quad
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insqh $17,$6,$4 # U : New bits
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bis $2,$4,$1 # E : Put it all together
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stq $1,0($5) # L : And back to memory
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ret $31,($26),1 # L0 :
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within_one_quad:
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ldq_u $1,0($16) # L :
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insql $17,$16,$2 # U : New bits
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mskql $1,$16,$4 # U : Clear old
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bis $2,$4,$2 # E : New result
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mskql $2,$6,$4 # U :
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mskqh $1,$6,$2 # U :
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bis $2,$4,$1 # E :
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stq_u $1,0($16) # L :
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end:
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nop
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nop
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nop
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ret $31,($26),1 # L0 :
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.end __constant_c_memset
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EXPORT_SYMBOL(__constant_c_memset)
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/*
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* This is a replicant of the __constant_c_memset code, rescheduled
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* to mask stalls. Note that entry point names also had to change
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*/
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.align 5
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.ent __memsetw
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__memsetw:
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.frame $30,0,$26,0
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.prologue 0
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inswl $17,0,$5 # U : 000000000000c1c2
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inswl $17,2,$2 # U : 00000000c1c20000
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bis $16,$16,$0 # E : return value
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addq $18,$16,$6 # E : max address to write to
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ble $18, end_w # U : zero length requested?
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inswl $17,4,$3 # U : 0000c1c200000000
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inswl $17,6,$4 # U : c1c2000000000000
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xor $16,$6,$1 # E : will complete write be within one quadword?
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or $2,$5,$2 # E : 00000000c1c2c1c2
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or $3,$4,$17 # E : c1c2c1c200000000
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bic $1,7,$1 # E : fit within a single quadword
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and $16,7,$3 # E : Target addr misalignment
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or $17,$2,$17 # E : c1c2c1c2c1c2c1c2
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beq $1,within_quad_w # U :
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nop
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beq $3,aligned_w # U : target is 0mod8
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/*
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* Target address is misaligned, and won't fit within a quadword
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*/
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ldq_u $4,0($16) # L : Fetch first partial
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bis $16,$16,$5 # E : Save the address
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insql $17,$16,$2 # U : Insert new bytes
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subq $3,8,$3 # E : Invert (for addressing uses)
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addq $18,$3,$18 # E : $18 is new count ($3 is negative)
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mskql $4,$16,$4 # U : clear relevant parts of the quad
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subq $16,$3,$16 # E : $16 is new aligned destination
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bis $2,$4,$1 # E : Final bytes
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nop
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stq_u $1,0($5) # L : Store result
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nop
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nop
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.align 4
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aligned_w:
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/*
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* We are now guaranteed to be quad aligned, with at least
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* one partial quad to write.
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*/
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sra $18,3,$3 # U : Number of remaining quads to write
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and $18,7,$18 # E : Number of trailing bytes to write
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bis $16,$16,$5 # E : Save dest address
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beq $3,no_quad_w # U : tail stuff only
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/*
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* it's worth the effort to unroll this and use wh64 if possible
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* Lifted a bunch of code from clear_user.S
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* At this point, entry values are:
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* $16 Current destination address
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* $5 A copy of $16
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* $6 The max quadword address to write to
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* $18 Number trailer bytes
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* $3 Number quads to write
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*/
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and $16, 0x3f, $2 # E : Forward work (only useful for unrolled loop)
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subq $3, 16, $4 # E : Only try to unroll if > 128 bytes
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subq $2, 0x40, $1 # E : bias counter (aligning stuff 0mod64)
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blt $4, loop_w # U :
|
|
|
|
/*
|
|
* We know we've got at least 16 quads, minimum of one trip
|
|
* through unrolled loop. Do a quad at a time to get us 0mod64
|
|
* aligned.
|
|
*/
|
|
|
|
nop # E :
|
|
nop # E :
|
|
nop # E :
|
|
beq $1, $bigalign_w # U :
|
|
|
|
$alignmod64_w:
|
|
stq $17, 0($5) # L :
|
|
subq $3, 1, $3 # E : For consistency later
|
|
addq $1, 8, $1 # E : Increment towards zero for alignment
|
|
addq $5, 8, $4 # E : Initial wh64 address (filler instruction)
|
|
|
|
nop
|
|
nop
|
|
addq $5, 8, $5 # E : Inc address
|
|
blt $1, $alignmod64_w # U :
|
|
|
|
$bigalign_w:
|
|
/*
|
|
* $3 - number quads left to go
|
|
* $5 - target address (aligned 0mod64)
|
|
* $17 - mask of stuff to store
|
|
* Scratch registers available: $7, $2, $4, $1
|
|
* we know that we'll be taking a minimum of one trip through
|
|
* CWG Section 3.7.6: do not expect a sustained store rate of > 1/cycle
|
|
* Assumes the wh64 needs to be for 2 trips through the loop in the future
|
|
* The wh64 is issued on for the starting destination address for trip +2
|
|
* through the loop, and if there are less than two trips left, the target
|
|
* address will be for the current trip.
|
|
*/
|
|
|
|
$do_wh64_w:
|
|
wh64 ($4) # L1 : memory subsystem write hint
|
|
subq $3, 24, $2 # E : For determining future wh64 addresses
|
|
stq $17, 0($5) # L :
|
|
nop # E :
|
|
|
|
addq $5, 128, $4 # E : speculative target of next wh64
|
|
stq $17, 8($5) # L :
|
|
stq $17, 16($5) # L :
|
|
addq $5, 64, $7 # E : Fallback address for wh64 (== next trip addr)
|
|
|
|
stq $17, 24($5) # L :
|
|
stq $17, 32($5) # L :
|
|
cmovlt $2, $7, $4 # E : Latency 2, extra mapping cycle
|
|
nop
|
|
|
|
stq $17, 40($5) # L :
|
|
stq $17, 48($5) # L :
|
|
subq $3, 16, $2 # E : Repeat the loop at least once more?
|
|
nop
|
|
|
|
stq $17, 56($5) # L :
|
|
addq $5, 64, $5 # E :
|
|
subq $3, 8, $3 # E :
|
|
bge $2, $do_wh64_w # U :
|
|
|
|
nop
|
|
nop
|
|
nop
|
|
beq $3, no_quad_w # U : Might have finished already
|
|
|
|
.align 4
|
|
/*
|
|
* Simple loop for trailing quadwords, or for small amounts
|
|
* of data (where we can't use an unrolled loop and wh64)
|
|
*/
|
|
loop_w:
|
|
stq $17,0($5) # L :
|
|
subq $3,1,$3 # E : Decrement number quads left
|
|
addq $5,8,$5 # E : Inc address
|
|
bne $3,loop_w # U : more?
|
|
|
|
no_quad_w:
|
|
/*
|
|
* Write 0..7 trailing bytes.
|
|
*/
|
|
nop # E :
|
|
beq $18,end_w # U : All done?
|
|
ldq $7,0($5) # L :
|
|
mskqh $7,$6,$2 # U : Mask final quad
|
|
|
|
insqh $17,$6,$4 # U : New bits
|
|
bis $2,$4,$1 # E : Put it all together
|
|
stq $1,0($5) # L : And back to memory
|
|
ret $31,($26),1 # L0 :
|
|
|
|
within_quad_w:
|
|
ldq_u $1,0($16) # L :
|
|
insql $17,$16,$2 # U : New bits
|
|
mskql $1,$16,$4 # U : Clear old
|
|
bis $2,$4,$2 # E : New result
|
|
|
|
mskql $2,$6,$4 # U :
|
|
mskqh $1,$6,$2 # U :
|
|
bis $2,$4,$1 # E :
|
|
stq_u $1,0($16) # L :
|
|
|
|
end_w:
|
|
nop
|
|
nop
|
|
nop
|
|
ret $31,($26),1 # L0 :
|
|
|
|
.end __memsetw
|
|
EXPORT_SYMBOL(__memsetw)
|
|
|
|
memset = ___memset
|
|
__memset = ___memset
|
|
EXPORT_SYMBOL(memset)
|
|
EXPORT_SYMBOL(__memset)
|