Adding upstream version 5.10.209.upstream/5.10.209upstream

Signed-off-by: Daniel Baumann <daniel.baumann@progress-linux.org>

1 files changed, 212 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..a28f39d34
--- /dev/null
+++ b/arch/ia64/lib/clear_user.S
@@ -0,0 +1,212 @@
+/* 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)