blob: eecd8577b2099c759a21ad486f24d61c831d951f [file] [log] [blame]
* 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 <>
#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).
.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, saved_lc
mov // preserve (slow)
;; // 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
;; p6,p0=16,len // if len > 16 then long memset
mov // initialize lc for small count
(p6) br.cond.dptk .long_do_clear
;; // WAR on
// 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.
EX( .Lexit1, st1 [buf]=r0,1 )
adds len=-1,len // countdown length using len
br.cloop.dptk 1b
;; // avoid RAW on
// .Lexit4: comes from byte by byte loop
// len contains bytes left
mov ret0=len // faster than using
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: 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 p6,p0=buf,1
EX( .Lexit3, (p6) st2 [buf]=r0,2 ) // 2-byte aligned
(p6) adds len=-2,len;; p6,p0=buf,2
EX( .Lexit3, (p6) st4 [buf]=r0,4 ) // 4-byte aligned
(p6) adds len=-4,len;; 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
// 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 and derive how many byte were left by
// doing : left= 16* + 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
// 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.
mov len2=len // for parallelization of error handling
mov len3=len p6,p0=len,3
EX( .Lexit2, (p6) st8 [buf]=r0,8 ) // at least 8 bytes
(p6) adds len3=-8,len2 p7,p6=len,2
EX( .Lexit2, (p7) st4 [buf]=r0,4 ) // at least 4 bytes
(p7) adds len2=-4,len3 p6,p7=len,1
EX( .Lexit2, (p6) st2 [buf]=r0,2 ) // at least 2 bytes
(p6) adds len3=-2,len2 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.
.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
mov ret0=len
br.ret.sptk.many rp