WIP FPC-III support

[linux/fpc-iii.git] / arch / ia64 / lib / clear_user.S

/* 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 <asm/asmmacro.h>
#include <asm/export.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)