Linux-2.6.12-rc2
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!
diff --git a/arch/ia64/lib/copy_user.S b/arch/ia64/lib/copy_user.S
new file mode 100644
index 0000000..c952bdc
--- /dev/null
+++ b/arch/ia64/lib/copy_user.S
@@ -0,0 +1,610 @@
+/*
+ *
+ * Optimized version of the copy_user() routine.
+ * It is used to copy date across the kernel/user boundary.
+ *
+ * The source and destination are always on opposite side of
+ * the boundary. When reading from user space we must catch
+ * faults on loads. When writing to user space we must catch
+ * errors on stores. Note that because of the nature of the copy
+ * we don't need to worry about overlapping regions.
+ *
+ *
+ * Inputs:
+ * in0 address of source buffer
+ * in1 address of destination buffer
+ * in2 number of bytes to copy
+ *
+ * Outputs:
+ * ret0 0 in case of success. The number of bytes NOT copied in
+ * case of error.
+ *
+ * Copyright (C) 2000-2001 Hewlett-Packard Co
+ * Stephane Eranian <eranian@hpl.hp.com>
+ *
+ * Fixme:
+ * - handle the case where we have more than 16 bytes and the alignment
+ * are different.
+ * - more benchmarking
+ * - fix extraneous stop bit introduced by the EX() macro.
+ */
+
+#include <asm/asmmacro.h>
+
+//
+// Tuneable parameters
+//
+#define COPY_BREAK 16 // we do byte copy below (must be >=16)
+#define PIPE_DEPTH 21 // pipe depth
+
+#define EPI p[PIPE_DEPTH-1]
+
+//
+// arguments
+//
+#define dst in0
+#define src in1
+#define len in2
+
+//
+// local registers
+//
+#define t1 r2 // rshift in bytes
+#define t2 r3 // lshift in bytes
+#define rshift r14 // right shift in bits
+#define lshift r15 // left shift in bits
+#define word1 r16
+#define word2 r17
+#define cnt r18
+#define len2 r19
+#define saved_lc r20
+#define saved_pr r21
+#define tmp r22
+#define val r23
+#define src1 r24
+#define dst1 r25
+#define src2 r26
+#define dst2 r27
+#define len1 r28
+#define enddst r29
+#define endsrc r30
+#define saved_pfs r31
+
+GLOBAL_ENTRY(__copy_user)
+ .prologue
+ .save ar.pfs, saved_pfs
+ alloc saved_pfs=ar.pfs,3,((2*PIPE_DEPTH+7)&~7),0,((2*PIPE_DEPTH+7)&~7)
+
+ .rotr val1[PIPE_DEPTH],val2[PIPE_DEPTH]
+ .rotp p[PIPE_DEPTH]
+
+ adds len2=-1,len // br.ctop is repeat/until
+ mov ret0=r0
+
+ ;; // RAW of cfm when len=0
+ cmp.eq p8,p0=r0,len // check for zero length
+ .save ar.lc, saved_lc
+ mov saved_lc=ar.lc // preserve ar.lc (slow)
+(p8) br.ret.spnt.many rp // empty mempcy()
+ ;;
+ add enddst=dst,len // first byte after end of source
+ add endsrc=src,len // first byte after end of destination
+ .save pr, saved_pr
+ mov saved_pr=pr // preserve predicates
+
+ .body
+
+ mov dst1=dst // copy because of rotation
+ mov ar.ec=PIPE_DEPTH
+ mov pr.rot=1<<16 // p16=true all others are false
+
+ mov src1=src // copy because of rotation
+ mov ar.lc=len2 // initialize lc for small count
+ cmp.lt p10,p7=COPY_BREAK,len // if len > COPY_BREAK then long copy
+
+ xor tmp=src,dst // same alignment test prepare
+(p10) br.cond.dptk .long_copy_user
+ ;; // RAW pr.rot/p16 ?
+ //
+ // Now we do the byte by byte loop with software pipeline
+ //
+ // p7 is necessarily false by now
+1:
+ EX(.failure_in_pipe1,(p16) ld1 val1[0]=[src1],1)
+ EX(.failure_out,(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1)
+ br.ctop.dptk.few 1b
+ ;;
+ mov ar.lc=saved_lc
+ mov pr=saved_pr,0xffffffffffff0000
+ mov ar.pfs=saved_pfs // restore ar.ec
+ br.ret.sptk.many rp // end of short memcpy
+
+ //
+ // Not 8-byte aligned
+ //
+.diff_align_copy_user:
+ // At this point we know we have more than 16 bytes to copy
+ // and also that src and dest do _not_ have the same alignment.
+ and src2=0x7,src1 // src offset
+ and dst2=0x7,dst1 // dst offset
+ ;;
+ // The basic idea is that we copy byte-by-byte at the head so
+ // that we can reach 8-byte alignment for both src1 and dst1.
+ // Then copy the body using software pipelined 8-byte copy,
+ // shifting the two back-to-back words right and left, then copy
+ // the tail by copying byte-by-byte.
+ //
+ // Fault handling. If the byte-by-byte at the head fails on the
+ // load, then restart and finish the pipleline by copying zeros
+ // to the dst1. Then copy zeros for the rest of dst1.
+ // If 8-byte software pipeline fails on the load, do the same as
+ // failure_in3 does. If the byte-by-byte at the tail fails, it is
+ // handled simply by failure_in_pipe1.
+ //
+ // The case p14 represents the source has more bytes in the
+ // the first word (by the shifted part), whereas the p15 needs to
+ // copy some bytes from the 2nd word of the source that has the
+ // tail of the 1st of the destination.
+ //
+
+ //
+ // Optimization. If dst1 is 8-byte aligned (quite common), we don't need
+ // to copy the head to dst1, to start 8-byte copy software pipeline.
+ // We know src1 is not 8-byte aligned in this case.
+ //
+ cmp.eq p14,p15=r0,dst2
+(p15) br.cond.spnt 1f
+ ;;
+ sub t1=8,src2
+ mov t2=src2
+ ;;
+ shl rshift=t2,3
+ sub len1=len,t1 // set len1
+ ;;
+ sub lshift=64,rshift
+ ;;
+ br.cond.spnt .word_copy_user
+ ;;
+1:
+ cmp.leu p14,p15=src2,dst2
+ sub t1=dst2,src2
+ ;;
+ .pred.rel "mutex", p14, p15
+(p14) sub word1=8,src2 // (8 - src offset)
+(p15) sub t1=r0,t1 // absolute value
+(p15) sub word1=8,dst2 // (8 - dst offset)
+ ;;
+ // For the case p14, we don't need to copy the shifted part to
+ // the 1st word of destination.
+ sub t2=8,t1
+(p14) sub word1=word1,t1
+ ;;
+ sub len1=len,word1 // resulting len
+(p15) shl rshift=t1,3 // in bits
+(p14) shl rshift=t2,3
+ ;;
+(p14) sub len1=len1,t1
+ adds cnt=-1,word1
+ ;;
+ sub lshift=64,rshift
+ mov ar.ec=PIPE_DEPTH
+ mov pr.rot=1<<16 // p16=true all others are false
+ mov ar.lc=cnt
+ ;;
+2:
+ EX(.failure_in_pipe2,(p16) ld1 val1[0]=[src1],1)
+ EX(.failure_out,(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1)
+ br.ctop.dptk.few 2b
+ ;;
+ clrrrb
+ ;;
+.word_copy_user:
+ cmp.gtu p9,p0=16,len1
+(p9) br.cond.spnt 4f // if (16 > len1) skip 8-byte copy
+ ;;
+ shr.u cnt=len1,3 // number of 64-bit words
+ ;;
+ adds cnt=-1,cnt
+ ;;
+ .pred.rel "mutex", p14, p15
+(p14) sub src1=src1,t2
+(p15) sub src1=src1,t1
+ //
+ // Now both src1 and dst1 point to an 8-byte aligned address. And
+ // we have more than 8 bytes to copy.
+ //
+ mov ar.lc=cnt
+ mov ar.ec=PIPE_DEPTH
+ mov pr.rot=1<<16 // p16=true all others are false
+ ;;
+3:
+ //
+ // The pipleline consists of 3 stages:
+ // 1 (p16): Load a word from src1
+ // 2 (EPI_1): Shift right pair, saving to tmp
+ // 3 (EPI): Store tmp to dst1
+ //
+ // To make it simple, use at least 2 (p16) loops to set up val1[n]
+ // because we need 2 back-to-back val1[] to get tmp.
+ // Note that this implies EPI_2 must be p18 or greater.
+ //
+
+#define EPI_1 p[PIPE_DEPTH-2]
+#define SWITCH(pred, shift) cmp.eq pred,p0=shift,rshift
+#define CASE(pred, shift) \
+ (pred) br.cond.spnt .copy_user_bit##shift
+#define BODY(rshift) \
+.copy_user_bit##rshift: \
+1: \
+ EX(.failure_out,(EPI) st8 [dst1]=tmp,8); \
+(EPI_1) shrp tmp=val1[PIPE_DEPTH-2],val1[PIPE_DEPTH-1],rshift; \
+ EX(3f,(p16) ld8 val1[1]=[src1],8); \
+(p16) mov val1[0]=r0; \
+ br.ctop.dptk 1b; \
+ ;; \
+ br.cond.sptk.many .diff_align_do_tail; \
+2: \
+(EPI) st8 [dst1]=tmp,8; \
+(EPI_1) shrp tmp=val1[PIPE_DEPTH-2],val1[PIPE_DEPTH-1],rshift; \
+3: \
+(p16) mov val1[1]=r0; \
+(p16) mov val1[0]=r0; \
+ br.ctop.dptk 2b; \
+ ;; \
+ br.cond.sptk.many .failure_in2
+
+ //
+ // Since the instruction 'shrp' requires a fixed 128-bit value
+ // specifying the bits to shift, we need to provide 7 cases
+ // below.
+ //
+ SWITCH(p6, 8)
+ SWITCH(p7, 16)
+ SWITCH(p8, 24)
+ SWITCH(p9, 32)
+ SWITCH(p10, 40)
+ SWITCH(p11, 48)
+ SWITCH(p12, 56)
+ ;;
+ CASE(p6, 8)
+ CASE(p7, 16)
+ CASE(p8, 24)
+ CASE(p9, 32)
+ CASE(p10, 40)
+ CASE(p11, 48)
+ CASE(p12, 56)
+ ;;
+ BODY(8)
+ BODY(16)
+ BODY(24)
+ BODY(32)
+ BODY(40)
+ BODY(48)
+ BODY(56)
+ ;;
+.diff_align_do_tail:
+ .pred.rel "mutex", p14, p15
+(p14) sub src1=src1,t1
+(p14) adds dst1=-8,dst1
+(p15) sub dst1=dst1,t1
+ ;;
+4:
+ // Tail correction.
+ //
+ // The problem with this piplelined loop is that the last word is not
+ // loaded and thus parf of the last word written is not correct.
+ // To fix that, we simply copy the tail byte by byte.
+
+ sub len1=endsrc,src1,1
+ clrrrb
+ ;;
+ mov ar.ec=PIPE_DEPTH
+ mov pr.rot=1<<16 // p16=true all others are false
+ mov ar.lc=len1
+ ;;
+5:
+ EX(.failure_in_pipe1,(p16) ld1 val1[0]=[src1],1)
+ EX(.failure_out,(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1)
+ br.ctop.dptk.few 5b
+ ;;
+ mov ar.lc=saved_lc
+ mov pr=saved_pr,0xffffffffffff0000
+ mov ar.pfs=saved_pfs
+ br.ret.sptk.many rp
+
+ //
+ // Beginning of long mempcy (i.e. > 16 bytes)
+ //
+.long_copy_user:
+ tbit.nz p6,p7=src1,0 // odd alignment
+ and tmp=7,tmp
+ ;;
+ cmp.eq p10,p8=r0,tmp
+ mov len1=len // copy because of rotation
+(p8) br.cond.dpnt .diff_align_copy_user
+ ;;
+ // At this point we know we have more than 16 bytes to copy
+ // and also that both src and dest have the same alignment
+ // which may not be the one we want. So for now we must move
+ // forward slowly until we reach 16byte alignment: no need to
+ // worry about reaching the end of buffer.
+ //
+ EX(.failure_in1,(p6) ld1 val1[0]=[src1],1) // 1-byte aligned
+(p6) adds len1=-1,len1;;
+ tbit.nz p7,p0=src1,1
+ ;;
+ EX(.failure_in1,(p7) ld2 val1[1]=[src1],2) // 2-byte aligned
+(p7) adds len1=-2,len1;;
+ tbit.nz p8,p0=src1,2
+ ;;
+ //
+ // Stop bit not required after ld4 because if we fail on ld4
+ // we have never executed the ld1, therefore st1 is not executed.
+ //
+ EX(.failure_in1,(p8) ld4 val2[0]=[src1],4) // 4-byte aligned
+ ;;
+ EX(.failure_out,(p6) st1 [dst1]=val1[0],1)
+ tbit.nz p9,p0=src1,3
+ ;;
+ //
+ // Stop bit not required after ld8 because if we fail on ld8
+ // we have never executed the ld2, therefore st2 is not executed.
+ //
+ EX(.failure_in1,(p9) ld8 val2[1]=[src1],8) // 8-byte aligned
+ EX(.failure_out,(p7) st2 [dst1]=val1[1],2)
+(p8) adds len1=-4,len1
+ ;;
+ EX(.failure_out, (p8) st4 [dst1]=val2[0],4)
+(p9) adds len1=-8,len1;;
+ shr.u cnt=len1,4 // number of 128-bit (2x64bit) words
+ ;;
+ EX(.failure_out, (p9) st8 [dst1]=val2[1],8)
+ tbit.nz p6,p0=len1,3
+ cmp.eq p7,p0=r0,cnt
+ adds tmp=-1,cnt // br.ctop is repeat/until
+(p7) br.cond.dpnt .dotail // we have less than 16 bytes left
+ ;;
+ adds src2=8,src1
+ adds dst2=8,dst1
+ mov ar.lc=tmp
+ ;;
+ //
+ // 16bytes/iteration
+ //
+2:
+ EX(.failure_in3,(p16) ld8 val1[0]=[src1],16)
+(p16) ld8 val2[0]=[src2],16
+
+ EX(.failure_out, (EPI) st8 [dst1]=val1[PIPE_DEPTH-1],16)
+(EPI) st8 [dst2]=val2[PIPE_DEPTH-1],16
+ br.ctop.dptk 2b
+ ;; // RAW on src1 when fall through from loop
+ //
+ // Tail correction based on len only
+ //
+ // No matter where we come from (loop or test) the src1 pointer
+ // is 16 byte aligned AND we have less than 16 bytes to copy.
+ //
+.dotail:
+ EX(.failure_in1,(p6) ld8 val1[0]=[src1],8) // at least 8 bytes
+ tbit.nz p7,p0=len1,2
+ ;;
+ EX(.failure_in1,(p7) ld4 val1[1]=[src1],4) // at least 4 bytes
+ tbit.nz p8,p0=len1,1
+ ;;
+ EX(.failure_in1,(p8) ld2 val2[0]=[src1],2) // at least 2 bytes
+ tbit.nz p9,p0=len1,0
+ ;;
+ EX(.failure_out, (p6) st8 [dst1]=val1[0],8)
+ ;;
+ EX(.failure_in1,(p9) ld1 val2[1]=[src1]) // only 1 byte left
+ mov ar.lc=saved_lc
+ ;;
+ EX(.failure_out,(p7) st4 [dst1]=val1[1],4)
+ mov pr=saved_pr,0xffffffffffff0000
+ ;;
+ EX(.failure_out, (p8) st2 [dst1]=val2[0],2)
+ mov ar.pfs=saved_pfs
+ ;;
+ EX(.failure_out, (p9) st1 [dst1]=val2[1])
+ br.ret.sptk.many rp
+
+
+ //
+ // Here we handle the case where the byte by byte copy fails
+ // on the load.
+ // Several factors make the zeroing of the rest of the buffer kind of
+ // tricky:
+ // - the pipeline: loads/stores are not in sync (pipeline)
+ //
+ // In the same loop iteration, the dst1 pointer does not directly
+ // reflect where the faulty load was.
+ //
+ // - pipeline effect
+ // When you get a fault on load, you may have valid data from
+ // previous loads not yet store in transit. Such data must be
+ // store normally before moving onto zeroing the rest.
+ //
+ // - single/multi dispersal independence.
+ //
+ // solution:
+ // - we don't disrupt the pipeline, i.e. data in transit in
+ // the software pipeline will be eventually move to memory.
+ // We simply replace the load with a simple mov and keep the
+ // pipeline going. We can't really do this inline because
+ // p16 is always reset to 1 when lc > 0.
+ //
+.failure_in_pipe1:
+ sub ret0=endsrc,src1 // number of bytes to zero, i.e. not copied
+1:
+(p16) mov val1[0]=r0
+(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1
+ br.ctop.dptk 1b
+ ;;
+ mov pr=saved_pr,0xffffffffffff0000
+ mov ar.lc=saved_lc
+ mov ar.pfs=saved_pfs
+ br.ret.sptk.many rp
+
+ //
+ // This is the case where the byte by byte copy fails on the load
+ // when we copy the head. We need to finish the pipeline and copy
+ // zeros for the rest of the destination. Since this happens
+ // at the top we still need to fill the body and tail.
+.failure_in_pipe2:
+ sub ret0=endsrc,src1 // number of bytes to zero, i.e. not copied
+2:
+(p16) mov val1[0]=r0
+(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1
+ br.ctop.dptk 2b
+ ;;
+ sub len=enddst,dst1,1 // precompute len
+ br.cond.dptk.many .failure_in1bis
+ ;;
+
+ //
+ // 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)