summaryrefslogtreecommitdiff
path: root/arch/ia64/lib/clear_user.S
diff options
context:
space:
mode:
authorLinus Torvalds <torvalds@ppc970.osdl.org>2005-04-16 15:20:36 -0700
committerLinus Torvalds <torvalds@ppc970.osdl.org>2005-04-16 15:20:36 -0700
commit1da177e4c3f41524e886b7f1b8a0c1fc7321cac2 (patch)
tree0bba044c4ce775e45a88a51686b5d9f90697ea9d /arch/ia64/lib/clear_user.S
downloadconfigs-2.6.12-rc2.tar.gz
Linux-2.6.12-rc2v2.6.12-rc2
Initial git repository build. I'm not bothering with the full history, even though we have it. We can create a separate "historical" git archive of that later if we want to, and in the meantime it's about 3.2GB when imported into git - space that would just make the early git days unnecessarily complicated, when we don't have a lot of good infrastructure for it. Let it rip!
Diffstat (limited to 'arch/ia64/lib/clear_user.S')
-rw-r--r--arch/ia64/lib/clear_user.S209
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 00000000000..eecd8577b20
--- /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)