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Initial git repository build. I'm not bothering with the full history, even though we have it. We can create a separate "historical" git archive of that later if we want to, and in the meantime it's about 3.2GB when imported into git - space that would just make the early git days unnecessarily complicated, when we don't have a lot of good infrastructure for it. Let it rip!
610 lines
17 KiB
ArmAsm
610 lines
17 KiB
ArmAsm
/*
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*
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* Optimized version of the copy_user() routine.
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* It is used to copy date across the kernel/user boundary.
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*
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* The source and destination are always on opposite side of
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* the boundary. When reading from user space we must catch
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* faults on loads. When writing to user space we must catch
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* errors on stores. Note that because of the nature of the copy
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* we don't need to worry about overlapping regions.
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*
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*
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* Inputs:
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* in0 address of source buffer
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* in1 address of destination buffer
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* in2 number of bytes to copy
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*
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* Outputs:
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* ret0 0 in case of success. The number of bytes NOT copied in
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* case of error.
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*
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* Copyright (C) 2000-2001 Hewlett-Packard Co
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* Stephane Eranian <eranian@hpl.hp.com>
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*
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* Fixme:
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* - handle the case where we have more than 16 bytes and the alignment
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* are different.
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* - more benchmarking
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* - fix extraneous stop bit introduced by the EX() macro.
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*/
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#include <asm/asmmacro.h>
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//
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// Tuneable parameters
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//
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#define COPY_BREAK 16 // we do byte copy below (must be >=16)
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#define PIPE_DEPTH 21 // pipe depth
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#define EPI p[PIPE_DEPTH-1]
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//
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// arguments
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//
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#define dst in0
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#define src in1
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#define len in2
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//
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// local registers
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//
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#define t1 r2 // rshift in bytes
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#define t2 r3 // lshift in bytes
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#define rshift r14 // right shift in bits
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#define lshift r15 // left shift in bits
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#define word1 r16
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#define word2 r17
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#define cnt r18
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#define len2 r19
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#define saved_lc r20
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#define saved_pr r21
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#define tmp r22
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#define val r23
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#define src1 r24
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#define dst1 r25
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#define src2 r26
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#define dst2 r27
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#define len1 r28
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#define enddst r29
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#define endsrc r30
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#define saved_pfs r31
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GLOBAL_ENTRY(__copy_user)
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.prologue
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.save ar.pfs, saved_pfs
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alloc saved_pfs=ar.pfs,3,((2*PIPE_DEPTH+7)&~7),0,((2*PIPE_DEPTH+7)&~7)
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.rotr val1[PIPE_DEPTH],val2[PIPE_DEPTH]
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.rotp p[PIPE_DEPTH]
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adds len2=-1,len // br.ctop is repeat/until
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mov ret0=r0
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;; // RAW of cfm when len=0
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cmp.eq p8,p0=r0,len // check for zero length
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.save ar.lc, saved_lc
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mov saved_lc=ar.lc // preserve ar.lc (slow)
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(p8) br.ret.spnt.many rp // empty mempcy()
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;;
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add enddst=dst,len // first byte after end of source
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add endsrc=src,len // first byte after end of destination
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.save pr, saved_pr
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mov saved_pr=pr // preserve predicates
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.body
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mov dst1=dst // copy because of rotation
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mov ar.ec=PIPE_DEPTH
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mov pr.rot=1<<16 // p16=true all others are false
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mov src1=src // copy because of rotation
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mov ar.lc=len2 // initialize lc for small count
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cmp.lt p10,p7=COPY_BREAK,len // if len > COPY_BREAK then long copy
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xor tmp=src,dst // same alignment test prepare
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(p10) br.cond.dptk .long_copy_user
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;; // RAW pr.rot/p16 ?
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//
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// Now we do the byte by byte loop with software pipeline
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//
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// p7 is necessarily false by now
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1:
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EX(.failure_in_pipe1,(p16) ld1 val1[0]=[src1],1)
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EX(.failure_out,(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1)
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br.ctop.dptk.few 1b
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;;
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mov ar.lc=saved_lc
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mov pr=saved_pr,0xffffffffffff0000
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mov ar.pfs=saved_pfs // restore ar.ec
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br.ret.sptk.many rp // end of short memcpy
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//
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// Not 8-byte aligned
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//
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.diff_align_copy_user:
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// At this point we know we have more than 16 bytes to copy
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// and also that src and dest do _not_ have the same alignment.
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and src2=0x7,src1 // src offset
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and dst2=0x7,dst1 // dst offset
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;;
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// The basic idea is that we copy byte-by-byte at the head so
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// that we can reach 8-byte alignment for both src1 and dst1.
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// Then copy the body using software pipelined 8-byte copy,
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// shifting the two back-to-back words right and left, then copy
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// the tail by copying byte-by-byte.
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//
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// Fault handling. If the byte-by-byte at the head fails on the
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// load, then restart and finish the pipleline by copying zeros
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// to the dst1. Then copy zeros for the rest of dst1.
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// If 8-byte software pipeline fails on the load, do the same as
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// failure_in3 does. If the byte-by-byte at the tail fails, it is
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// handled simply by failure_in_pipe1.
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//
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// The case p14 represents the source has more bytes in the
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// the first word (by the shifted part), whereas the p15 needs to
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// copy some bytes from the 2nd word of the source that has the
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// tail of the 1st of the destination.
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//
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//
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// Optimization. If dst1 is 8-byte aligned (quite common), we don't need
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// to copy the head to dst1, to start 8-byte copy software pipeline.
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// We know src1 is not 8-byte aligned in this case.
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//
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cmp.eq p14,p15=r0,dst2
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(p15) br.cond.spnt 1f
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;;
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sub t1=8,src2
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mov t2=src2
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;;
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shl rshift=t2,3
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sub len1=len,t1 // set len1
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;;
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sub lshift=64,rshift
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;;
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br.cond.spnt .word_copy_user
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;;
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1:
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cmp.leu p14,p15=src2,dst2
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sub t1=dst2,src2
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;;
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.pred.rel "mutex", p14, p15
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(p14) sub word1=8,src2 // (8 - src offset)
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(p15) sub t1=r0,t1 // absolute value
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(p15) sub word1=8,dst2 // (8 - dst offset)
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;;
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// For the case p14, we don't need to copy the shifted part to
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// the 1st word of destination.
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sub t2=8,t1
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(p14) sub word1=word1,t1
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;;
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sub len1=len,word1 // resulting len
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(p15) shl rshift=t1,3 // in bits
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(p14) shl rshift=t2,3
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;;
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(p14) sub len1=len1,t1
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adds cnt=-1,word1
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;;
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sub lshift=64,rshift
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mov ar.ec=PIPE_DEPTH
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mov pr.rot=1<<16 // p16=true all others are false
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mov ar.lc=cnt
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;;
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2:
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EX(.failure_in_pipe2,(p16) ld1 val1[0]=[src1],1)
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EX(.failure_out,(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1)
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br.ctop.dptk.few 2b
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;;
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clrrrb
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;;
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.word_copy_user:
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cmp.gtu p9,p0=16,len1
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(p9) br.cond.spnt 4f // if (16 > len1) skip 8-byte copy
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;;
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shr.u cnt=len1,3 // number of 64-bit words
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;;
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adds cnt=-1,cnt
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;;
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.pred.rel "mutex", p14, p15
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(p14) sub src1=src1,t2
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(p15) sub src1=src1,t1
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//
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// Now both src1 and dst1 point to an 8-byte aligned address. And
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// we have more than 8 bytes to copy.
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//
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mov ar.lc=cnt
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mov ar.ec=PIPE_DEPTH
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mov pr.rot=1<<16 // p16=true all others are false
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;;
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3:
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//
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// The pipleline consists of 3 stages:
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// 1 (p16): Load a word from src1
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// 2 (EPI_1): Shift right pair, saving to tmp
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// 3 (EPI): Store tmp to dst1
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//
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// To make it simple, use at least 2 (p16) loops to set up val1[n]
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// because we need 2 back-to-back val1[] to get tmp.
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// Note that this implies EPI_2 must be p18 or greater.
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//
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#define EPI_1 p[PIPE_DEPTH-2]
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#define SWITCH(pred, shift) cmp.eq pred,p0=shift,rshift
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#define CASE(pred, shift) \
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(pred) br.cond.spnt .copy_user_bit##shift
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#define BODY(rshift) \
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.copy_user_bit##rshift: \
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1: \
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EX(.failure_out,(EPI) st8 [dst1]=tmp,8); \
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(EPI_1) shrp tmp=val1[PIPE_DEPTH-2],val1[PIPE_DEPTH-1],rshift; \
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EX(3f,(p16) ld8 val1[1]=[src1],8); \
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(p16) mov val1[0]=r0; \
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br.ctop.dptk 1b; \
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;; \
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br.cond.sptk.many .diff_align_do_tail; \
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2: \
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(EPI) st8 [dst1]=tmp,8; \
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(EPI_1) shrp tmp=val1[PIPE_DEPTH-2],val1[PIPE_DEPTH-1],rshift; \
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3: \
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(p16) mov val1[1]=r0; \
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(p16) mov val1[0]=r0; \
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br.ctop.dptk 2b; \
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;; \
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br.cond.sptk.many .failure_in2
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//
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// Since the instruction 'shrp' requires a fixed 128-bit value
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// specifying the bits to shift, we need to provide 7 cases
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// below.
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//
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SWITCH(p6, 8)
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SWITCH(p7, 16)
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SWITCH(p8, 24)
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SWITCH(p9, 32)
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SWITCH(p10, 40)
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SWITCH(p11, 48)
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SWITCH(p12, 56)
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;;
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CASE(p6, 8)
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CASE(p7, 16)
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CASE(p8, 24)
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CASE(p9, 32)
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CASE(p10, 40)
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CASE(p11, 48)
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CASE(p12, 56)
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;;
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BODY(8)
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BODY(16)
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BODY(24)
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BODY(32)
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BODY(40)
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BODY(48)
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BODY(56)
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;;
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.diff_align_do_tail:
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.pred.rel "mutex", p14, p15
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(p14) sub src1=src1,t1
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(p14) adds dst1=-8,dst1
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(p15) sub dst1=dst1,t1
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;;
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4:
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// Tail correction.
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//
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// The problem with this piplelined loop is that the last word is not
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// loaded and thus parf of the last word written is not correct.
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// To fix that, we simply copy the tail byte by byte.
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sub len1=endsrc,src1,1
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clrrrb
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;;
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mov ar.ec=PIPE_DEPTH
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mov pr.rot=1<<16 // p16=true all others are false
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mov ar.lc=len1
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;;
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5:
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EX(.failure_in_pipe1,(p16) ld1 val1[0]=[src1],1)
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EX(.failure_out,(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1)
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br.ctop.dptk.few 5b
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;;
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mov ar.lc=saved_lc
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mov pr=saved_pr,0xffffffffffff0000
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mov ar.pfs=saved_pfs
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br.ret.sptk.many rp
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//
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// Beginning of long mempcy (i.e. > 16 bytes)
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//
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.long_copy_user:
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tbit.nz p6,p7=src1,0 // odd alignment
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and tmp=7,tmp
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;;
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cmp.eq p10,p8=r0,tmp
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mov len1=len // copy because of rotation
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(p8) br.cond.dpnt .diff_align_copy_user
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;;
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// At this point we know we have more than 16 bytes to copy
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// and also that both src and dest have the same alignment
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// which may not be the one we want. So for now we must move
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// forward slowly until we reach 16byte alignment: no need to
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// worry about reaching the end of buffer.
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//
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EX(.failure_in1,(p6) ld1 val1[0]=[src1],1) // 1-byte aligned
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(p6) adds len1=-1,len1;;
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tbit.nz p7,p0=src1,1
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;;
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EX(.failure_in1,(p7) ld2 val1[1]=[src1],2) // 2-byte aligned
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(p7) adds len1=-2,len1;;
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tbit.nz p8,p0=src1,2
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;;
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//
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// Stop bit not required after ld4 because if we fail on ld4
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// we have never executed the ld1, therefore st1 is not executed.
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//
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EX(.failure_in1,(p8) ld4 val2[0]=[src1],4) // 4-byte aligned
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;;
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EX(.failure_out,(p6) st1 [dst1]=val1[0],1)
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tbit.nz p9,p0=src1,3
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;;
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//
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// Stop bit not required after ld8 because if we fail on ld8
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// we have never executed the ld2, therefore st2 is not executed.
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//
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EX(.failure_in1,(p9) ld8 val2[1]=[src1],8) // 8-byte aligned
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EX(.failure_out,(p7) st2 [dst1]=val1[1],2)
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(p8) adds len1=-4,len1
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;;
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EX(.failure_out, (p8) st4 [dst1]=val2[0],4)
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(p9) adds len1=-8,len1;;
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shr.u cnt=len1,4 // number of 128-bit (2x64bit) words
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;;
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EX(.failure_out, (p9) st8 [dst1]=val2[1],8)
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tbit.nz p6,p0=len1,3
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cmp.eq p7,p0=r0,cnt
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adds tmp=-1,cnt // br.ctop is repeat/until
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(p7) br.cond.dpnt .dotail // we have less than 16 bytes left
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;;
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adds src2=8,src1
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adds dst2=8,dst1
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mov ar.lc=tmp
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;;
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//
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// 16bytes/iteration
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//
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2:
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EX(.failure_in3,(p16) ld8 val1[0]=[src1],16)
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(p16) ld8 val2[0]=[src2],16
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EX(.failure_out, (EPI) st8 [dst1]=val1[PIPE_DEPTH-1],16)
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(EPI) st8 [dst2]=val2[PIPE_DEPTH-1],16
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br.ctop.dptk 2b
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;; // RAW on src1 when fall through from loop
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//
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// Tail correction based on len only
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//
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// No matter where we come from (loop or test) the src1 pointer
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// is 16 byte aligned AND we have less than 16 bytes to copy.
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//
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.dotail:
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EX(.failure_in1,(p6) ld8 val1[0]=[src1],8) // at least 8 bytes
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tbit.nz p7,p0=len1,2
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;;
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EX(.failure_in1,(p7) ld4 val1[1]=[src1],4) // at least 4 bytes
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tbit.nz p8,p0=len1,1
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;;
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EX(.failure_in1,(p8) ld2 val2[0]=[src1],2) // at least 2 bytes
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tbit.nz p9,p0=len1,0
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;;
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EX(.failure_out, (p6) st8 [dst1]=val1[0],8)
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;;
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EX(.failure_in1,(p9) ld1 val2[1]=[src1]) // only 1 byte left
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mov ar.lc=saved_lc
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;;
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EX(.failure_out,(p7) st4 [dst1]=val1[1],4)
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mov pr=saved_pr,0xffffffffffff0000
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;;
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EX(.failure_out, (p8) st2 [dst1]=val2[0],2)
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mov ar.pfs=saved_pfs
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;;
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EX(.failure_out, (p9) st1 [dst1]=val2[1])
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br.ret.sptk.many rp
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//
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// Here we handle the case where the byte by byte copy fails
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// on the load.
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// Several factors make the zeroing of the rest of the buffer kind of
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// tricky:
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// - the pipeline: loads/stores are not in sync (pipeline)
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//
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// In the same loop iteration, the dst1 pointer does not directly
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// reflect where the faulty load was.
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//
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// - pipeline effect
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// When you get a fault on load, you may have valid data from
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// previous loads not yet store in transit. Such data must be
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// store normally before moving onto zeroing the rest.
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//
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// - single/multi dispersal independence.
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//
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// solution:
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// - we don't disrupt the pipeline, i.e. data in transit in
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// the software pipeline will be eventually move to memory.
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// We simply replace the load with a simple mov and keep the
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// pipeline going. We can't really do this inline because
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// p16 is always reset to 1 when lc > 0.
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//
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.failure_in_pipe1:
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sub ret0=endsrc,src1 // number of bytes to zero, i.e. not copied
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1:
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(p16) mov val1[0]=r0
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(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1
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br.ctop.dptk 1b
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;;
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mov pr=saved_pr,0xffffffffffff0000
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mov ar.lc=saved_lc
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mov ar.pfs=saved_pfs
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br.ret.sptk.many rp
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//
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// This is the case where the byte by byte copy fails on the load
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// when we copy the head. We need to finish the pipeline and copy
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// zeros for the rest of the destination. Since this happens
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// at the top we still need to fill the body and tail.
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.failure_in_pipe2:
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sub ret0=endsrc,src1 // number of bytes to zero, i.e. not copied
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2:
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(p16) mov val1[0]=r0
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(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1
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br.ctop.dptk 2b
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;;
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sub len=enddst,dst1,1 // precompute len
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br.cond.dptk.many .failure_in1bis
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;;
|
|
|
|
//
|
|
// Here we handle the head & tail part when we check for alignment.
|
|
// The following code handles only the load failures. The
|
|
// main diffculty comes from the fact that loads/stores are
|
|
// scheduled. So when you fail on a load, the stores corresponding
|
|
// to previous successful loads must be executed.
|
|
//
|
|
// However some simplifications are possible given the way
|
|
// things work.
|
|
//
|
|
// 1) HEAD
|
|
// Theory of operation:
|
|
//
|
|
// Page A | Page B
|
|
// ---------|-----
|
|
// 1|8 x
|
|
// 1 2|8 x
|
|
// 4|8 x
|
|
// 1 4|8 x
|
|
// 2 4|8 x
|
|
// 1 2 4|8 x
|
|
// |1
|
|
// |2 x
|
|
// |4 x
|
|
//
|
|
// page_size >= 4k (2^12). (x means 4, 2, 1)
|
|
// Here we suppose Page A exists and Page B does not.
|
|
//
|
|
// As we move towards eight byte alignment we may encounter faults.
|
|
// The numbers on each page show the size of the load (current alignment).
|
|
//
|
|
// Key point:
|
|
// - if you fail on 1, 2, 4 then you have never executed any smaller
|
|
// size loads, e.g. failing ld4 means no ld1 nor ld2 executed
|
|
// before.
|
|
//
|
|
// This allows us to simplify the cleanup code, because basically you
|
|
// only have to worry about "pending" stores in the case of a failing
|
|
// ld8(). Given the way the code is written today, this means only
|
|
// worry about st2, st4. There we can use the information encapsulated
|
|
// into the predicates.
|
|
//
|
|
// Other key point:
|
|
// - if you fail on the ld8 in the head, it means you went straight
|
|
// to it, i.e. 8byte alignment within an unexisting page.
|
|
// Again this comes from the fact that if you crossed just for the ld8 then
|
|
// you are 8byte aligned but also 16byte align, therefore you would
|
|
// either go for the 16byte copy loop OR the ld8 in the tail part.
|
|
// The combination ld1, ld2, ld4, ld8 where you fail on ld8 is impossible
|
|
// because it would mean you had 15bytes to copy in which case you
|
|
// would have defaulted to the byte by byte copy.
|
|
//
|
|
//
|
|
// 2) TAIL
|
|
// Here we now we have less than 16 bytes AND we are either 8 or 16 byte
|
|
// aligned.
|
|
//
|
|
// Key point:
|
|
// This means that we either:
|
|
// - are right on a page boundary
|
|
// OR
|
|
// - are at more than 16 bytes from a page boundary with
|
|
// at most 15 bytes to copy: no chance of crossing.
|
|
//
|
|
// This allows us to assume that if we fail on a load we haven't possibly
|
|
// executed any of the previous (tail) ones, so we don't need to do
|
|
// any stores. For instance, if we fail on ld2, this means we had
|
|
// 2 or 3 bytes left to copy and we did not execute the ld8 nor ld4.
|
|
//
|
|
// This means that we are in a situation similar the a fault in the
|
|
// head part. That's nice!
|
|
//
|
|
.failure_in1:
|
|
sub ret0=endsrc,src1 // number of bytes to zero, i.e. not copied
|
|
sub len=endsrc,src1,1
|
|
//
|
|
// we know that ret0 can never be zero at this point
|
|
// because we failed why trying to do a load, i.e. there is still
|
|
// some work to do.
|
|
// The failure_in1bis and length problem is taken care of at the
|
|
// calling side.
|
|
//
|
|
;;
|
|
.failure_in1bis: // from (.failure_in3)
|
|
mov ar.lc=len // Continue with a stupid byte store.
|
|
;;
|
|
5:
|
|
st1 [dst1]=r0,1
|
|
br.cloop.dptk 5b
|
|
;;
|
|
mov pr=saved_pr,0xffffffffffff0000
|
|
mov ar.lc=saved_lc
|
|
mov ar.pfs=saved_pfs
|
|
br.ret.sptk.many rp
|
|
|
|
//
|
|
// Here we simply restart the loop but instead
|
|
// of doing loads we fill the pipeline with zeroes
|
|
// We can't simply store r0 because we may have valid
|
|
// data in transit in the pipeline.
|
|
// ar.lc and ar.ec are setup correctly at this point
|
|
//
|
|
// we MUST use src1/endsrc here and not dst1/enddst because
|
|
// of the pipeline effect.
|
|
//
|
|
.failure_in3:
|
|
sub ret0=endsrc,src1 // number of bytes to zero, i.e. not copied
|
|
;;
|
|
2:
|
|
(p16) mov val1[0]=r0
|
|
(p16) mov val2[0]=r0
|
|
(EPI) st8 [dst1]=val1[PIPE_DEPTH-1],16
|
|
(EPI) st8 [dst2]=val2[PIPE_DEPTH-1],16
|
|
br.ctop.dptk 2b
|
|
;;
|
|
cmp.ne p6,p0=dst1,enddst // Do we need to finish the tail ?
|
|
sub len=enddst,dst1,1 // precompute len
|
|
(p6) br.cond.dptk .failure_in1bis
|
|
;;
|
|
mov pr=saved_pr,0xffffffffffff0000
|
|
mov ar.lc=saved_lc
|
|
mov ar.pfs=saved_pfs
|
|
br.ret.sptk.many rp
|
|
|
|
.failure_in2:
|
|
sub ret0=endsrc,src1
|
|
cmp.ne p6,p0=dst1,enddst // Do we need to finish the tail ?
|
|
sub len=enddst,dst1,1 // precompute len
|
|
(p6) br.cond.dptk .failure_in1bis
|
|
;;
|
|
mov pr=saved_pr,0xffffffffffff0000
|
|
mov ar.lc=saved_lc
|
|
mov ar.pfs=saved_pfs
|
|
br.ret.sptk.many rp
|
|
|
|
//
|
|
// handling of failures on stores: that's the easy part
|
|
//
|
|
.failure_out:
|
|
sub ret0=enddst,dst1
|
|
mov pr=saved_pr,0xffffffffffff0000
|
|
mov ar.lc=saved_lc
|
|
|
|
mov ar.pfs=saved_pfs
|
|
br.ret.sptk.many rp
|
|
END(__copy_user)
|