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Diffstat (limited to 'kernel/arch/ia64/lib/copy_user.S')
| -rw-r--r-- | kernel/arch/ia64/lib/copy_user.S | 610 |
1 files changed, 610 insertions, 0 deletions
diff --git a/kernel/arch/ia64/lib/copy_user.S b/kernel/arch/ia64/lib/copy_user.S new file mode 100644 index 000000000..c952bdc6a --- /dev/null +++ b/kernel/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) |