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Diffstat (limited to 'arch/ia64/lib/clear_user.S')
| -rw-r--r-- | arch/ia64/lib/clear_user.S | 212 |
1 files changed, 0 insertions, 212 deletions
diff --git a/arch/ia64/lib/clear_user.S b/arch/ia64/lib/clear_user.S deleted file mode 100644 index 1d9e45ccf8..0000000000 --- a/arch/ia64/lib/clear_user.S +++ /dev/null @@ -1,212 +0,0 @@ -/* SPDX-License-Identifier: GPL-2.0 */ -/* - * 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 <linux/export.h> -#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) -EXPORT_SYMBOL(__do_clear_user) |