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259 lines
7.3 KiB
259 lines
7.3 KiB
/*
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* arch/alpha/lib/ev6-copy_user.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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* Copy to/from user space, handling exceptions as we go.. This
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* isn't exactly pretty.
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*
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* This is essentially the same as "memcpy()", but with a few twists.
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* Notably, we have to make sure that $0 is always up-to-date and
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* contains the right "bytes left to copy" value (and that it is updated
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* only _after_ a successful copy). There is also some rather minor
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* exception setup stuff..
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*
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* NOTE! This is not directly C-callable, because the calling semantics are
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* different:
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*
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* Inputs:
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* length in $0
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* destination address in $6
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* source address in $7
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* return address in $28
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*
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* Outputs:
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* bytes left to copy in $0
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*
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* Clobbers:
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* $1,$2,$3,$4,$5,$6,$7
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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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*/
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/* Allow an exception for an insn; exit if we get one. */
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#define EXI(x,y...) \
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99: x,##y; \
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.section __ex_table,"a"; \
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.long 99b - .; \
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lda $31, $exitin-99b($31); \
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.previous
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#define EXO(x,y...) \
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99: x,##y; \
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.section __ex_table,"a"; \
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.long 99b - .; \
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lda $31, $exitout-99b($31); \
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.previous
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.set noat
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.align 4
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.globl __copy_user
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.ent __copy_user
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# Pipeline info: Slotting & Comments
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__copy_user:
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.prologue 0
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subq $0, 32, $1 # .. E .. .. : Is this going to be a small copy?
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beq $0, $zerolength # U .. .. .. : U L U L
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and $6,7,$3 # .. .. .. E : is leading dest misalignment
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ble $1, $onebyteloop # .. .. U .. : 1st branch : small amount of data
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beq $3, $destaligned # .. U .. .. : 2nd (one cycle fetcher stall)
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subq $3, 8, $3 # E .. .. .. : L U U L : trip counter
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/*
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* The fetcher stall also hides the 1 cycle cross-cluster stall for $3 (L --> U)
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* This loop aligns the destination a byte at a time
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* We know we have at least one trip through this loop
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*/
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$aligndest:
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EXI( ldbu $1,0($7) ) # .. .. .. L : Keep loads separate from stores
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addq $6,1,$6 # .. .. E .. : Section 3.8 in the CWG
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addq $3,1,$3 # .. E .. .. :
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nop # E .. .. .. : U L U L
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/*
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* the -1 is to compensate for the inc($6) done in a previous quadpack
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* which allows us zero dependencies within either quadpack in the loop
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*/
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EXO( stb $1,-1($6) ) # .. .. .. L :
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addq $7,1,$7 # .. .. E .. : Section 3.8 in the CWG
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subq $0,1,$0 # .. E .. .. :
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bne $3, $aligndest # U .. .. .. : U L U L
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/*
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* If we fell through into here, we have a minimum of 33 - 7 bytes
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* If we arrived via branch, we have a minimum of 32 bytes
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*/
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$destaligned:
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and $7,7,$1 # .. .. .. E : Check _current_ source alignment
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bic $0,7,$4 # .. .. E .. : number bytes as a quadword loop
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EXI( ldq_u $3,0($7) ) # .. L .. .. : Forward fetch for fallthrough code
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beq $1,$quadaligned # U .. .. .. : U L U L
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/*
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* In the worst case, we've just executed an ldq_u here from 0($7)
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* and we'll repeat it once if we take the branch
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*/
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/* Misaligned quadword loop - not unrolled. Leave it that way. */
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$misquad:
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EXI( ldq_u $2,8($7) ) # .. .. .. L :
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subq $4,8,$4 # .. .. E .. :
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extql $3,$7,$3 # .. U .. .. :
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extqh $2,$7,$1 # U .. .. .. : U U L L
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bis $3,$1,$1 # .. .. .. E :
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EXO( stq $1,0($6) ) # .. .. L .. :
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addq $7,8,$7 # .. E .. .. :
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subq $0,8,$0 # E .. .. .. : U L L U
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addq $6,8,$6 # .. .. .. E :
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bis $2,$2,$3 # .. .. E .. :
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nop # .. E .. .. :
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bne $4,$misquad # U .. .. .. : U L U L
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nop # .. .. .. E
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nop # .. .. E ..
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nop # .. E .. ..
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beq $0,$zerolength # U .. .. .. : U L U L
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/* We know we have at least one trip through the byte loop */
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EXI ( ldbu $2,0($7) ) # .. .. .. L : No loads in the same quad
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addq $6,1,$6 # .. .. E .. : as the store (Section 3.8 in CWG)
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nop # .. E .. .. :
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br $31, $dirtyentry # L0 .. .. .. : L U U L
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/* Do the trailing byte loop load, then hop into the store part of the loop */
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/*
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* A minimum of (33 - 7) bytes to do a quad at a time.
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* Based upon the usage context, it's worth the effort to unroll this loop
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* $0 - number of bytes to be moved
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* $4 - number of bytes to move as quadwords
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* $6 is current destination address
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* $7 is current source address
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*/
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$quadaligned:
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subq $4, 32, $2 # .. .. .. E : do not unroll for small stuff
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nop # .. .. E ..
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nop # .. E .. ..
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blt $2, $onequad # U .. .. .. : U L U L
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/*
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* There is a significant assumption here that the source and destination
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* addresses differ by more than 32 bytes. In this particular case, a
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* sparsity of registers further bounds this to be a minimum of 8 bytes.
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* But if this isn't met, then the output result will be incorrect.
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* Furthermore, due to a lack of available registers, we really can't
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* unroll this to be an 8x loop (which would enable us to use the wh64
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* instruction memory hint instruction).
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*/
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$unroll4:
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EXI( ldq $1,0($7) ) # .. .. .. L
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EXI( ldq $2,8($7) ) # .. .. L ..
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subq $4,32,$4 # .. E .. ..
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nop # E .. .. .. : U U L L
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addq $7,16,$7 # .. .. .. E
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EXO( stq $1,0($6) ) # .. .. L ..
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EXO( stq $2,8($6) ) # .. L .. ..
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subq $0,16,$0 # E .. .. .. : U L L U
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addq $6,16,$6 # .. .. .. E
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EXI( ldq $1,0($7) ) # .. .. L ..
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EXI( ldq $2,8($7) ) # .. L .. ..
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subq $4, 32, $3 # E .. .. .. : U U L L : is there enough for another trip?
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EXO( stq $1,0($6) ) # .. .. .. L
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EXO( stq $2,8($6) ) # .. .. L ..
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subq $0,16,$0 # .. E .. ..
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addq $7,16,$7 # E .. .. .. : U L L U
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nop # .. .. .. E
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nop # .. .. E ..
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addq $6,16,$6 # .. E .. ..
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bgt $3,$unroll4 # U .. .. .. : U L U L
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nop
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nop
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nop
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beq $4, $noquads
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$onequad:
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EXI( ldq $1,0($7) )
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subq $4,8,$4
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addq $7,8,$7
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nop
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EXO( stq $1,0($6) )
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subq $0,8,$0
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addq $6,8,$6
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bne $4,$onequad
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$noquads:
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nop
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nop
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nop
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beq $0,$zerolength
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/*
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* For small copies (or the tail of a larger copy), do a very simple byte loop.
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* There's no point in doing a lot of complex alignment calculations to try to
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* to quadword stuff for a small amount of data.
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* $0 - remaining number of bytes left to copy
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* $6 - current dest addr
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* $7 - current source addr
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*/
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$onebyteloop:
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EXI ( ldbu $2,0($7) ) # .. .. .. L : No loads in the same quad
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addq $6,1,$6 # .. .. E .. : as the store (Section 3.8 in CWG)
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nop # .. E .. .. :
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nop # E .. .. .. : U L U L
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$dirtyentry:
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/*
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* the -1 is to compensate for the inc($6) done in a previous quadpack
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* which allows us zero dependencies within either quadpack in the loop
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*/
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EXO ( stb $2,-1($6) ) # .. .. .. L :
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addq $7,1,$7 # .. .. E .. : quadpack as the load
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subq $0,1,$0 # .. E .. .. : change count _after_ copy
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bgt $0,$onebyteloop # U .. .. .. : U L U L
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$zerolength:
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$exitout: # Destination for exception recovery(?)
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nop # .. .. .. E
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nop # .. .. E ..
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nop # .. E .. ..
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ret $31,($28),1 # L0 .. .. .. : L U L U
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$exitin:
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/* A stupid byte-by-byte zeroing of the rest of the output
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buffer. This cures security holes by never leaving
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random kernel data around to be copied elsewhere. */
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nop
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nop
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nop
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mov $0,$1
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$101:
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EXO ( stb $31,0($6) ) # L
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subq $1,1,$1 # E
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addq $6,1,$6 # E
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bgt $1,$101 # U
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nop
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nop
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nop
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ret $31,($28),1 # L0
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.end __copy_user
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