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Diffstat (limited to 'arch/ia64/lib/clear_user.S')
| -rw-r--r-- | arch/ia64/lib/clear_user.S | 209 |
1 files changed, 209 insertions, 0 deletions
diff --git a/arch/ia64/lib/clear_user.S b/arch/ia64/lib/clear_user.S new file mode 100644 index 000000000..eecd8577b --- /dev/null +++ b/arch/ia64/lib/clear_user.S @@ -0,0 +1,209 @@ +/* + * This routine clears to zero a linear memory buffer in user space. + * + * Inputs: + * in0: address of buffer + * in1: length of buffer in bytes + * Outputs: + * r8: number of bytes that didn't get cleared due to a fault + * + * Copyright (C) 1998, 1999, 2001 Hewlett-Packard Co + * Stephane Eranian <eranian@hpl.hp.com> + */ + +#include <asm/asmmacro.h> + +// +// arguments +// +#define buf r32 +#define len r33 + +// +// local registers +// +#define cnt r16 +#define buf2 r17 +#define saved_lc r18 +#define saved_pfs r19 +#define tmp r20 +#define len2 r21 +#define len3 r22 + +// +// Theory of operations: +// - we check whether or not the buffer is small, i.e., less than 17 +// in which case we do the byte by byte loop. +// +// - Otherwise we go progressively from 1 byte store to 8byte store in +// the head part, the body is a 16byte store loop and we finish we the +// tail for the last 15 bytes. +// The good point about this breakdown is that the long buffer handling +// contains only 2 branches. +// +// The reason for not using shifting & masking for both the head and the +// tail is to stay semantically correct. This routine is not supposed +// to write bytes outside of the buffer. While most of the time this would +// be ok, we can't tolerate a mistake. A classical example is the case +// of multithreaded code were to the extra bytes touched is actually owned +// by another thread which runs concurrently to ours. Another, less likely, +// example is with device drivers where reading an I/O mapped location may +// have side effects (same thing for writing). +// + +GLOBAL_ENTRY(__do_clear_user) + .prologue + .save ar.pfs, saved_pfs + alloc saved_pfs=ar.pfs,2,0,0,0 + cmp.eq p6,p0=r0,len // check for zero length + .save ar.lc, saved_lc + mov saved_lc=ar.lc // preserve ar.lc (slow) + .body + ;; // avoid WAW on CFM + adds tmp=-1,len // br.ctop is repeat/until + mov ret0=len // return value is length at this point +(p6) br.ret.spnt.many rp + ;; + cmp.lt p6,p0=16,len // if len > 16 then long memset + mov ar.lc=tmp // initialize lc for small count +(p6) br.cond.dptk .long_do_clear + ;; // WAR on ar.lc + // + // worst case 16 iterations, avg 8 iterations + // + // We could have played with the predicates to use the extra + // M slot for 2 stores/iteration but the cost the initialization + // the various counters compared to how long the loop is supposed + // to last on average does not make this solution viable. + // +1: + EX( .Lexit1, st1 [buf]=r0,1 ) + adds len=-1,len // countdown length using len + br.cloop.dptk 1b + ;; // avoid RAW on ar.lc + // + // .Lexit4: comes from byte by byte loop + // len contains bytes left +.Lexit1: + mov ret0=len // faster than using ar.lc + mov ar.lc=saved_lc + br.ret.sptk.many rp // end of short clear_user + + + // + // At this point we know we have more than 16 bytes to copy + // so we focus on alignment (no branches required) + // + // The use of len/len2 for countdown of the number of bytes left + // instead of ret0 is due to the fact that the exception code + // changes the values of r8. + // +.long_do_clear: + tbit.nz p6,p0=buf,0 // odd alignment (for long_do_clear) + ;; + EX( .Lexit3, (p6) st1 [buf]=r0,1 ) // 1-byte aligned +(p6) adds len=-1,len;; // sync because buf is modified + tbit.nz p6,p0=buf,1 + ;; + EX( .Lexit3, (p6) st2 [buf]=r0,2 ) // 2-byte aligned +(p6) adds len=-2,len;; + tbit.nz p6,p0=buf,2 + ;; + EX( .Lexit3, (p6) st4 [buf]=r0,4 ) // 4-byte aligned +(p6) adds len=-4,len;; + tbit.nz p6,p0=buf,3 + ;; + EX( .Lexit3, (p6) st8 [buf]=r0,8 ) // 8-byte aligned +(p6) adds len=-8,len;; + shr.u cnt=len,4 // number of 128-bit (2x64bit) words + ;; + cmp.eq p6,p0=r0,cnt + adds tmp=-1,cnt +(p6) br.cond.dpnt .dotail // we have less than 16 bytes left + ;; + adds buf2=8,buf // setup second base pointer + mov ar.lc=tmp + ;; + + // + // 16bytes/iteration core loop + // + // The second store can never generate a fault because + // we come into the loop only when we are 16-byte aligned. + // This means that if we cross a page then it will always be + // in the first store and never in the second. + // + // + // We need to keep track of the remaining length. A possible (optimistic) + // way would be to use ar.lc and derive how many byte were left by + // doing : left= 16*ar.lc + 16. this would avoid the addition at + // every iteration. + // However we need to keep the synchronization point. A template + // M;;MB does not exist and thus we can keep the addition at no + // extra cycle cost (use a nop slot anyway). It also simplifies the + // (unlikely) error recovery code + // + +2: EX(.Lexit3, st8 [buf]=r0,16 ) + ;; // needed to get len correct when error + st8 [buf2]=r0,16 + adds len=-16,len + br.cloop.dptk 2b + ;; + mov ar.lc=saved_lc + // + // tail correction based on len only + // + // We alternate the use of len3,len2 to allow parallelism and correct + // error handling. We also reuse p6/p7 to return correct value. + // The addition of len2/len3 does not cost anything more compared to + // the regular memset as we had empty slots. + // +.dotail: + mov len2=len // for parallelization of error handling + mov len3=len + tbit.nz p6,p0=len,3 + ;; + EX( .Lexit2, (p6) st8 [buf]=r0,8 ) // at least 8 bytes +(p6) adds len3=-8,len2 + tbit.nz p7,p6=len,2 + ;; + EX( .Lexit2, (p7) st4 [buf]=r0,4 ) // at least 4 bytes +(p7) adds len2=-4,len3 + tbit.nz p6,p7=len,1 + ;; + EX( .Lexit2, (p6) st2 [buf]=r0,2 ) // at least 2 bytes +(p6) adds len3=-2,len2 + tbit.nz p7,p6=len,0 + ;; + EX( .Lexit2, (p7) st1 [buf]=r0 ) // only 1 byte left + mov ret0=r0 // success + br.ret.sptk.many rp // end of most likely path + + // + // Outlined error handling code + // + + // + // .Lexit3: comes from core loop, need restore pr/lc + // len contains bytes left + // + // + // .Lexit2: + // if p6 -> coming from st8 or st2 : len2 contains what's left + // if p7 -> coming from st4 or st1 : len3 contains what's left + // We must restore lc/pr even though might not have been used. +.Lexit2: + .pred.rel "mutex", p6, p7 +(p6) mov len=len2 +(p7) mov len=len3 + ;; + // + // .Lexit4: comes from head, need not restore pr/lc + // len contains bytes left + // +.Lexit3: + mov ret0=len + mov ar.lc=saved_lc + br.ret.sptk.many rp +END(__do_clear_user) |