3 * Optimized version of the copy_user() routine.
4 * It is used to copy date across the kernel/user boundary.
6 * The source and destination are always on opposite side of
7 * the boundary. When reading from user space we must catch
8 * faults on loads. When writing to user space we must catch
9 * errors on stores. Note that because of the nature of the copy
10 * we don't need to worry about overlapping regions.
14 * in0 address of source buffer
15 * in1 address of destination buffer
16 * in2 number of bytes to copy
19 * ret0 0 in case of success. The number of bytes NOT copied in
22 * Copyright (C) 2000-2001 Hewlett-Packard Co
23 * Stephane Eranian <eranian@hpl.hp.com>
26 * - handle the case where we have more than 16 bytes and the alignment
29 * - fix extraneous stop bit introduced by the EX() macro.
32 #include <asm/asmmacro.h>
35 // Tuneable parameters
37 #define COPY_BREAK 16 // we do byte copy below (must be >=16)
38 #define PIPE_DEPTH 21 // pipe depth
40 #define EPI p[PIPE_DEPTH-1]
52 #define t1 r2 // rshift in bytes
53 #define t2 r3 // lshift in bytes
54 #define rshift r14 // right shift in bits
55 #define lshift r15 // left shift in bits
73 GLOBAL_ENTRY(__copy_user)
75 .save ar.pfs, saved_pfs
76 alloc saved_pfs=ar.pfs,3,((2*PIPE_DEPTH+7)&~7),0,((2*PIPE_DEPTH+7)&~7)
78 .rotr val1[PIPE_DEPTH],val2[PIPE_DEPTH]
81 adds len2=-1,len // br.ctop is repeat/until
84 ;; // RAW of cfm when len=0
85 cmp.eq p8,p0=r0,len // check for zero length
87 mov saved_lc=ar.lc // preserve ar.lc (slow)
88 (p8) br.ret.spnt.many rp // empty mempcy()
90 add enddst=dst,len // first byte after end of source
91 add endsrc=src,len // first byte after end of destination
93 mov saved_pr=pr // preserve predicates
97 mov dst1=dst // copy because of rotation
99 mov pr.rot=1<<16 // p16=true all others are false
101 mov src1=src // copy because of rotation
102 mov ar.lc=len2 // initialize lc for small count
103 cmp.lt p10,p7=COPY_BREAK,len // if len > COPY_BREAK then long copy
105 xor tmp=src,dst // same alignment test prepare
106 (p10) br.cond.dptk .long_copy_user
107 ;; // RAW pr.rot/p16 ?
109 // Now we do the byte by byte loop with software pipeline
111 // p7 is necessarily false by now
113 EX(.failure_in_pipe1,(p16) ld1 val1[0]=[src1],1)
114 EX(.failure_out,(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1)
118 mov pr=saved_pr,0xffffffffffff0000
119 mov ar.pfs=saved_pfs // restore ar.ec
120 br.ret.sptk.many rp // end of short memcpy
123 // Not 8-byte aligned
125 .diff_align_copy_user:
126 // At this point we know we have more than 16 bytes to copy
127 // and also that src and dest do _not_ have the same alignment.
128 and src2=0x7,src1 // src offset
129 and dst2=0x7,dst1 // dst offset
131 // The basic idea is that we copy byte-by-byte at the head so
132 // that we can reach 8-byte alignment for both src1 and dst1.
133 // Then copy the body using software pipelined 8-byte copy,
134 // shifting the two back-to-back words right and left, then copy
135 // the tail by copying byte-by-byte.
137 // Fault handling. If the byte-by-byte at the head fails on the
138 // load, then restart and finish the pipleline by copying zeros
139 // to the dst1. Then copy zeros for the rest of dst1.
140 // If 8-byte software pipeline fails on the load, do the same as
141 // failure_in3 does. If the byte-by-byte at the tail fails, it is
142 // handled simply by failure_in_pipe1.
144 // The case p14 represents the source has more bytes in the
145 // the first word (by the shifted part), whereas the p15 needs to
146 // copy some bytes from the 2nd word of the source that has the
147 // tail of the 1st of the destination.
151 // Optimization. If dst1 is 8-byte aligned (quite common), we don't need
152 // to copy the head to dst1, to start 8-byte copy software pipeline.
153 // We know src1 is not 8-byte aligned in this case.
155 cmp.eq p14,p15=r0,dst2
156 (p15) br.cond.spnt 1f
162 sub len1=len,t1 // set len1
166 br.cond.spnt .word_copy_user
169 cmp.leu p14,p15=src2,dst2
172 .pred.rel "mutex", p14, p15
173 (p14) sub word1=8,src2 // (8 - src offset)
174 (p15) sub t1=r0,t1 // absolute value
175 (p15) sub word1=8,dst2 // (8 - dst offset)
177 // For the case p14, we don't need to copy the shifted part to
178 // the 1st word of destination.
180 (p14) sub word1=word1,t1
182 sub len1=len,word1 // resulting len
183 (p15) shl rshift=t1,3 // in bits
184 (p14) shl rshift=t2,3
186 (p14) sub len1=len1,t1
191 mov pr.rot=1<<16 // p16=true all others are false
195 EX(.failure_in_pipe2,(p16) ld1 val1[0]=[src1],1)
196 EX(.failure_out,(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1)
202 cmp.gtu p9,p0=16,len1
203 (p9) br.cond.spnt 4f // if (16 > len1) skip 8-byte copy
205 shr.u cnt=len1,3 // number of 64-bit words
209 .pred.rel "mutex", p14, p15
210 (p14) sub src1=src1,t2
211 (p15) sub src1=src1,t1
213 // Now both src1 and dst1 point to an 8-byte aligned address. And
214 // we have more than 8 bytes to copy.
218 mov pr.rot=1<<16 // p16=true all others are false
222 // The pipleline consists of 3 stages:
223 // 1 (p16): Load a word from src1
224 // 2 (EPI_1): Shift right pair, saving to tmp
225 // 3 (EPI): Store tmp to dst1
227 // To make it simple, use at least 2 (p16) loops to set up val1[n]
228 // because we need 2 back-to-back val1[] to get tmp.
229 // Note that this implies EPI_2 must be p18 or greater.
232 #define EPI_1 p[PIPE_DEPTH-2]
233 #define SWITCH(pred, shift) cmp.eq pred,p0=shift,rshift
234 #define CASE(pred, shift) \
235 (pred) br.cond.spnt .copy_user_bit##shift
236 #define BODY(rshift) \
237 .copy_user_bit##rshift: \
239 EX(.failure_out,(EPI) st8 [dst1]=tmp,8); \
240 (EPI_1) shrp tmp=val1[PIPE_DEPTH-2],val1[PIPE_DEPTH-1],rshift; \
241 EX(3f,(p16) ld8 val1[1]=[src1],8); \
242 (p16) mov val1[0]=r0; \
245 br.cond.sptk.many .diff_align_do_tail; \
247 (EPI) st8 [dst1]=tmp,8; \
248 (EPI_1) shrp tmp=val1[PIPE_DEPTH-2],val1[PIPE_DEPTH-1],rshift; \
250 (p16) mov val1[1]=r0; \
251 (p16) mov val1[0]=r0; \
254 br.cond.sptk.many .failure_in2
257 // Since the instruction 'shrp' requires a fixed 128-bit value
258 // specifying the bits to shift, we need to provide 7 cases
286 .pred.rel "mutex", p14, p15
287 (p14) sub src1=src1,t1
288 (p14) adds dst1=-8,dst1
289 (p15) sub dst1=dst1,t1
294 // The problem with this piplelined loop is that the last word is not
295 // loaded and thus parf of the last word written is not correct.
296 // To fix that, we simply copy the tail byte by byte.
298 sub len1=endsrc,src1,1
302 mov pr.rot=1<<16 // p16=true all others are false
306 EX(.failure_in_pipe1,(p16) ld1 val1[0]=[src1],1)
307 EX(.failure_out,(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1)
311 mov pr=saved_pr,0xffffffffffff0000
316 // Beginning of long mempcy (i.e. > 16 bytes)
319 tbit.nz p6,p7=src1,0 // odd alignment
323 mov len1=len // copy because of rotation
324 (p8) br.cond.dpnt .diff_align_copy_user
326 // At this point we know we have more than 16 bytes to copy
327 // and also that both src and dest have the same alignment
328 // which may not be the one we want. So for now we must move
329 // forward slowly until we reach 16byte alignment: no need to
330 // worry about reaching the end of buffer.
332 EX(.failure_in1,(p6) ld1 val1[0]=[src1],1) // 1-byte aligned
333 (p6) adds len1=-1,len1;;
336 EX(.failure_in1,(p7) ld2 val1[1]=[src1],2) // 2-byte aligned
337 (p7) adds len1=-2,len1;;
341 // Stop bit not required after ld4 because if we fail on ld4
342 // we have never executed the ld1, therefore st1 is not executed.
344 EX(.failure_in1,(p8) ld4 val2[0]=[src1],4) // 4-byte aligned
346 EX(.failure_out,(p6) st1 [dst1]=val1[0],1)
350 // Stop bit not required after ld8 because if we fail on ld8
351 // we have never executed the ld2, therefore st2 is not executed.
353 EX(.failure_in1,(p9) ld8 val2[1]=[src1],8) // 8-byte aligned
354 EX(.failure_out,(p7) st2 [dst1]=val1[1],2)
355 (p8) adds len1=-4,len1
357 EX(.failure_out, (p8) st4 [dst1]=val2[0],4)
358 (p9) adds len1=-8,len1;;
359 shr.u cnt=len1,4 // number of 128-bit (2x64bit) words
361 EX(.failure_out, (p9) st8 [dst1]=val2[1],8)
364 adds tmp=-1,cnt // br.ctop is repeat/until
365 (p7) br.cond.dpnt .dotail // we have less than 16 bytes left
375 EX(.failure_in3,(p16) ld8 val1[0]=[src1],16)
376 (p16) ld8 val2[0]=[src2],16
378 EX(.failure_out, (EPI) st8 [dst1]=val1[PIPE_DEPTH-1],16)
379 (EPI) st8 [dst2]=val2[PIPE_DEPTH-1],16
381 ;; // RAW on src1 when fall through from loop
383 // Tail correction based on len only
385 // No matter where we come from (loop or test) the src1 pointer
386 // is 16 byte aligned AND we have less than 16 bytes to copy.
389 EX(.failure_in1,(p6) ld8 val1[0]=[src1],8) // at least 8 bytes
392 EX(.failure_in1,(p7) ld4 val1[1]=[src1],4) // at least 4 bytes
395 EX(.failure_in1,(p8) ld2 val2[0]=[src1],2) // at least 2 bytes
398 EX(.failure_out, (p6) st8 [dst1]=val1[0],8)
400 EX(.failure_in1,(p9) ld1 val2[1]=[src1]) // only 1 byte left
403 EX(.failure_out,(p7) st4 [dst1]=val1[1],4)
404 mov pr=saved_pr,0xffffffffffff0000
406 EX(.failure_out, (p8) st2 [dst1]=val2[0],2)
409 EX(.failure_out, (p9) st1 [dst1]=val2[1])
414 // Here we handle the case where the byte by byte copy fails
416 // Several factors make the zeroing of the rest of the buffer kind of
418 // - the pipeline: loads/stores are not in sync (pipeline)
420 // In the same loop iteration, the dst1 pointer does not directly
421 // reflect where the faulty load was.
424 // When you get a fault on load, you may have valid data from
425 // previous loads not yet store in transit. Such data must be
426 // store normally before moving onto zeroing the rest.
428 // - single/multi dispersal independence.
431 // - we don't disrupt the pipeline, i.e. data in transit in
432 // the software pipeline will be eventually move to memory.
433 // We simply replace the load with a simple mov and keep the
434 // pipeline going. We can't really do this inline because
435 // p16 is always reset to 1 when lc > 0.
438 sub ret0=endsrc,src1 // number of bytes to zero, i.e. not copied
441 (EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1
444 mov pr=saved_pr,0xffffffffffff0000
450 // This is the case where the byte by byte copy fails on the load
451 // when we copy the head. We need to finish the pipeline and copy
452 // zeros for the rest of the destination. Since this happens
453 // at the top we still need to fill the body and tail.
455 sub ret0=endsrc,src1 // number of bytes to zero, i.e. not copied
458 (EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1
461 sub len=enddst,dst1,1 // precompute len
462 br.cond.dptk.many .failure_in1bis
466 // Here we handle the head & tail part when we check for alignment.
467 // The following code handles only the load failures. The
468 // main diffculty comes from the fact that loads/stores are
469 // scheduled. So when you fail on a load, the stores corresponding
470 // to previous successful loads must be executed.
472 // However some simplifications are possible given the way
476 // Theory of operation:
490 // page_size >= 4k (2^12). (x means 4, 2, 1)
491 // Here we suppose Page A exists and Page B does not.
493 // As we move towards eight byte alignment we may encounter faults.
494 // The numbers on each page show the size of the load (current alignment).
497 // - if you fail on 1, 2, 4 then you have never executed any smaller
498 // size loads, e.g. failing ld4 means no ld1 nor ld2 executed
501 // This allows us to simplify the cleanup code, because basically you
502 // only have to worry about "pending" stores in the case of a failing
503 // ld8(). Given the way the code is written today, this means only
504 // worry about st2, st4. There we can use the information encapsulated
505 // into the predicates.
508 // - if you fail on the ld8 in the head, it means you went straight
509 // to it, i.e. 8byte alignment within an unexisting page.
510 // Again this comes from the fact that if you crossed just for the ld8 then
511 // you are 8byte aligned but also 16byte align, therefore you would
512 // either go for the 16byte copy loop OR the ld8 in the tail part.
513 // The combination ld1, ld2, ld4, ld8 where you fail on ld8 is impossible
514 // because it would mean you had 15bytes to copy in which case you
515 // would have defaulted to the byte by byte copy.
519 // Here we now we have less than 16 bytes AND we are either 8 or 16 byte
523 // This means that we either:
524 // - are right on a page boundary
526 // - are at more than 16 bytes from a page boundary with
527 // at most 15 bytes to copy: no chance of crossing.
529 // This allows us to assume that if we fail on a load we haven't possibly
530 // executed any of the previous (tail) ones, so we don't need to do
531 // any stores. For instance, if we fail on ld2, this means we had
532 // 2 or 3 bytes left to copy and we did not execute the ld8 nor ld4.
534 // This means that we are in a situation similar the a fault in the
535 // head part. That's nice!
538 sub ret0=endsrc,src1 // number of bytes to zero, i.e. not copied
539 sub len=endsrc,src1,1
541 // we know that ret0 can never be zero at this point
542 // because we failed why trying to do a load, i.e. there is still
544 // The failure_in1bis and length problem is taken care of at the
548 .failure_in1bis: // from (.failure_in3)
549 mov ar.lc=len // Continue with a stupid byte store.
555 mov pr=saved_pr,0xffffffffffff0000
561 // Here we simply restart the loop but instead
562 // of doing loads we fill the pipeline with zeroes
563 // We can't simply store r0 because we may have valid
564 // data in transit in the pipeline.
565 // ar.lc and ar.ec are setup correctly at this point
567 // we MUST use src1/endsrc here and not dst1/enddst because
568 // of the pipeline effect.
571 sub ret0=endsrc,src1 // number of bytes to zero, i.e. not copied
576 (EPI) st8 [dst1]=val1[PIPE_DEPTH-1],16
577 (EPI) st8 [dst2]=val2[PIPE_DEPTH-1],16
580 cmp.ne p6,p0=dst1,enddst // Do we need to finish the tail ?
581 sub len=enddst,dst1,1 // precompute len
582 (p6) br.cond.dptk .failure_in1bis
584 mov pr=saved_pr,0xffffffffffff0000
591 cmp.ne p6,p0=dst1,enddst // Do we need to finish the tail ?
592 sub len=enddst,dst1,1 // precompute len
593 (p6) br.cond.dptk .failure_in1bis
595 mov pr=saved_pr,0xffffffffffff0000
601 // handling of failures on stores: that's the easy part
605 mov pr=saved_pr,0xffffffffffff0000