Adding upstream version 6.6.15.upstream/6.6.15

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

1 files changed, 613 insertions, 0 deletions

diff --git a/arch/ia64/lib/copy_user.S b/arch/ia64/lib/copy_user.S
new file mode 100644
index 0000000000..8daab72cfe
--- /dev/null
+++ b/arch/ia64/lib/copy_user.S
@@ -0,0 +1,613 @@
+/* SPDX-License-Identifier: GPL-2.0 */
+/*
+ *
+ * Optimized version of the copy_user() routine.
+ * It is used to copy date across the kernel/user boundary.
+ *
+ * The source and destination are always on opposite side of
+ * the boundary. When reading from user space we must catch
+ * faults on loads. When writing to user space we must catch
+ * errors on stores. Note that because of the nature of the copy
+ * we don't need to worry about overlapping regions.
+ *
+ *
+ * Inputs:
+ *	in0	address of source buffer
+ *	in1	address of destination buffer
+ *	in2	number of bytes to copy
+ *
+ * Outputs:
+ *	ret0	0 in case of success. The number of bytes NOT copied in
+ *		case of error.
+ *
+ * Copyright (C) 2000-2001 Hewlett-Packard Co
+ *	Stephane Eranian <eranian@hpl.hp.com>
+ *
+ * Fixme:
+ *	- handle the case where we have more than 16 bytes and the alignment
+ *	  are different.
+ *	- more benchmarking
+ *	- fix extraneous stop bit introduced by the EX() macro.
+ */
+
+#include <linux/export.h>
+#include <asm/asmmacro.h>
+
+//
+// Tuneable parameters
+//
+#define COPY_BREAK	16	// we do byte copy below (must be >=16)
+#define PIPE_DEPTH	21	// pipe depth
+
+#define EPI		p[PIPE_DEPTH-1]
+
+//
+// arguments
+//
+#define dst		in0
+#define src		in1
+#define len		in2
+
+//
+// local registers
+//
+#define t1		r2	// rshift in bytes
+#define t2		r3	// lshift in bytes
+#define rshift		r14	// right shift in bits
+#define lshift		r15	// left shift in bits
+#define word1		r16
+#define word2		r17
+#define cnt		r18
+#define len2		r19
+#define saved_lc	r20
+#define saved_pr	r21
+#define tmp		r22
+#define val		r23
+#define src1		r24
+#define dst1		r25
+#define src2		r26
+#define dst2		r27
+#define len1		r28
+#define enddst		r29
+#define endsrc		r30
+#define saved_pfs	r31
+
+GLOBAL_ENTRY(__copy_user)
+	.prologue
+	.save ar.pfs, saved_pfs
+	alloc saved_pfs=ar.pfs,3,((2*PIPE_DEPTH+7)&~7),0,((2*PIPE_DEPTH+7)&~7)
+
+	.rotr val1[PIPE_DEPTH],val2[PIPE_DEPTH]
+	.rotp p[PIPE_DEPTH]
+
+	adds len2=-1,len	// br.ctop is repeat/until
+	mov ret0=r0
+
+	;;			// RAW of cfm when len=0
+	cmp.eq p8,p0=r0,len	// check for zero length
+	.save ar.lc, saved_lc
+	mov saved_lc=ar.lc	// preserve ar.lc (slow)
+(p8)	br.ret.spnt.many rp	// empty mempcy()
+	;;
+	add enddst=dst,len	// first byte after end of source
+	add endsrc=src,len	// first byte after end of destination
+	.save pr, saved_pr
+	mov saved_pr=pr		// preserve predicates
+
+	.body
+
+	mov dst1=dst		// copy because of rotation
+	mov ar.ec=PIPE_DEPTH
+	mov pr.rot=1<<16	// p16=true all others are false
+
+	mov src1=src		// copy because of rotation
+	mov ar.lc=len2		// initialize lc for small count
+	cmp.lt p10,p7=COPY_BREAK,len	// if len > COPY_BREAK then long copy
+
+	xor tmp=src,dst		// same alignment test prepare
+(p10)	br.cond.dptk .long_copy_user
+	;;			// RAW pr.rot/p16 ?
+	//
+	// Now we do the byte by byte loop with software pipeline
+	//
+	// p7 is necessarily false by now
+1:
+	EX(.failure_in_pipe1,(p16) ld1 val1[0]=[src1],1)
+	EX(.failure_out,(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1)
+	br.ctop.dptk.few 1b
+	;;
+	mov ar.lc=saved_lc
+	mov pr=saved_pr,0xffffffffffff0000
+	mov ar.pfs=saved_pfs		// restore ar.ec
+	br.ret.sptk.many rp		// end of short memcpy
+
+	//
+	// Not 8-byte aligned
+	//
+.diff_align_copy_user:
+	// At this point we know we have more than 16 bytes to copy
+	// and also that src and dest do _not_ have the same alignment.
+	and src2=0x7,src1				// src offset
+	and dst2=0x7,dst1				// dst offset
+	;;
+	// The basic idea is that we copy byte-by-byte at the head so
+	// that we can reach 8-byte alignment for both src1 and dst1.
+	// Then copy the body using software pipelined 8-byte copy,
+	// shifting the two back-to-back words right and left, then copy
+	// the tail by copying byte-by-byte.
+	//
+	// Fault handling. If the byte-by-byte at the head fails on the
+	// load, then restart and finish the pipleline by copying zeros
+	// to the dst1. Then copy zeros for the rest of dst1.
+	// If 8-byte software pipeline fails on the load, do the same as
+	// failure_in3 does. If the byte-by-byte at the tail fails, it is
+	// handled simply by failure_in_pipe1.
+	//
+	// The case p14 represents the source has more bytes in the
+	// the first word (by the shifted part), whereas the p15 needs to
+	// copy some bytes from the 2nd word of the source that has the
+	// tail of the 1st of the destination.
+	//
+
+	//
+	// Optimization. If dst1 is 8-byte aligned (quite common), we don't need
+	// to copy the head to dst1, to start 8-byte copy software pipeline.
+	// We know src1 is not 8-byte aligned in this case.
+	//
+	cmp.eq p14,p15=r0,dst2
+(p15)	br.cond.spnt 1f
+	;;
+	sub t1=8,src2
+	mov t2=src2
+	;;
+	shl rshift=t2,3
+	sub len1=len,t1					// set len1
+	;;
+	sub lshift=64,rshift
+	;;
+	br.cond.spnt .word_copy_user
+	;;
+1:
+	cmp.leu	p14,p15=src2,dst2
+	sub t1=dst2,src2
+	;;
+	.pred.rel "mutex", p14, p15
+(p14)	sub word1=8,src2				// (8 - src offset)
+(p15)	sub t1=r0,t1					// absolute value
+(p15)	sub word1=8,dst2				// (8 - dst offset)
+	;;
+	// For the case p14, we don't need to copy the shifted part to
+	// the 1st word of destination.
+	sub t2=8,t1
+(p14)	sub word1=word1,t1
+	;;
+	sub len1=len,word1				// resulting len
+(p15)	shl rshift=t1,3					// in bits
+(p14)	shl rshift=t2,3
+	;;
+(p14)	sub len1=len1,t1
+	adds cnt=-1,word1
+	;;
+	sub lshift=64,rshift
+	mov ar.ec=PIPE_DEPTH
+	mov pr.rot=1<<16	// p16=true all others are false
+	mov ar.lc=cnt
+	;;
+2:
+	EX(.failure_in_pipe2,(p16) ld1 val1[0]=[src1],1)
+	EX(.failure_out,(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1)
+	br.ctop.dptk.few 2b
+	;;
+	clrrrb
+	;;
+.word_copy_user:
+	cmp.gtu p9,p0=16,len1
+(p9)	br.cond.spnt 4f			// if (16 > len1) skip 8-byte copy
+	;;
+	shr.u cnt=len1,3		// number of 64-bit words
+	;;
+	adds cnt=-1,cnt
+	;;
+	.pred.rel "mutex", p14, p15
+(p14)	sub src1=src1,t2
+(p15)	sub src1=src1,t1
+	//
+	// Now both src1 and dst1 point to an 8-byte aligned address. And
+	// we have more than 8 bytes to copy.
+	//
+	mov ar.lc=cnt
+	mov ar.ec=PIPE_DEPTH
+	mov pr.rot=1<<16	// p16=true all others are false
+	;;
+3:
+	//
+	// The pipleline consists of 3 stages:
+	// 1 (p16):	Load a word from src1
+	// 2 (EPI_1):	Shift right pair, saving to tmp
+	// 3 (EPI):	Store tmp to dst1
+	//
+	// To make it simple, use at least 2 (p16) loops to set up val1[n]
+	// because we need 2 back-to-back val1[] to get tmp.
+	// Note that this implies EPI_2 must be p18 or greater.
+	//
+
+#define EPI_1		p[PIPE_DEPTH-2]
+#define SWITCH(pred, shift)	cmp.eq pred,p0=shift,rshift
+#define CASE(pred, shift)	\
+	(pred)	br.cond.spnt .copy_user_bit##shift
+#define BODY(rshift)						\
+.copy_user_bit##rshift:						\
+1:								\
+	EX(.failure_out,(EPI) st8 [dst1]=tmp,8);		\
+(EPI_1) shrp tmp=val1[PIPE_DEPTH-2],val1[PIPE_DEPTH-1],rshift;	\
+	EX(3f,(p16) ld8 val1[1]=[src1],8);			\
+(p16)	mov val1[0]=r0;						\
+	br.ctop.dptk 1b;					\
+	;;							\
+	br.cond.sptk.many .diff_align_do_tail;			\
+2:								\
+(EPI)	st8 [dst1]=tmp,8;					\
+(EPI_1)	shrp tmp=val1[PIPE_DEPTH-2],val1[PIPE_DEPTH-1],rshift;	\
+3:								\
+(p16)	mov val1[1]=r0;						\
+(p16)	mov val1[0]=r0;						\
+	br.ctop.dptk 2b;					\
+	;;							\
+	br.cond.sptk.many .failure_in2
+
+	//
+	// Since the instruction 'shrp' requires a fixed 128-bit value
+	// specifying the bits to shift, we need to provide 7 cases
+	// below.
+	//
+	SWITCH(p6, 8)
+	SWITCH(p7, 16)
+	SWITCH(p8, 24)
+	SWITCH(p9, 32)
+	SWITCH(p10, 40)
+	SWITCH(p11, 48)
+	SWITCH(p12, 56)
+	;;
+	CASE(p6, 8)
+	CASE(p7, 16)
+	CASE(p8, 24)
+	CASE(p9, 32)
+	CASE(p10, 40)
+	CASE(p11, 48)
+	CASE(p12, 56)
+	;;
+	BODY(8)
+	BODY(16)
+	BODY(24)
+	BODY(32)
+	BODY(40)
+	BODY(48)
+	BODY(56)
+	;;
+.diff_align_do_tail:
+	.pred.rel "mutex", p14, p15
+(p14)	sub src1=src1,t1
+(p14)	adds dst1=-8,dst1
+(p15)	sub dst1=dst1,t1
+	;;
+4:
+	// Tail correction.
+	//
+	// The problem with this piplelined loop is that the last word is not
+	// loaded and thus parf of the last word written is not correct.
+	// To fix that, we simply copy the tail byte by byte.
+
+	sub len1=endsrc,src1,1
+	clrrrb
+	;;
+	mov ar.ec=PIPE_DEPTH
+	mov pr.rot=1<<16	// p16=true all others are false
+	mov ar.lc=len1
+	;;
+5:
+	EX(.failure_in_pipe1,(p16) ld1 val1[0]=[src1],1)
+	EX(.failure_out,(EPI) st1 [dst1]=val1[PIPE_DEPTH-1],1)
+	br.ctop.dptk.few 5b
+	;;
+	mov ar.lc=saved_lc
+	mov pr=saved_pr,0xffffffffffff0000
+	mov ar.pfs=saved_pfs
+	br.ret.sptk.many rp
+
+	//
+	// Beginning of long mempcy (i.e. > 16 bytes)
+	//
+.long_copy_user:
+	tbit.nz p6,p7=src1,0	// odd alignment
+	and tmp=7,tmp
+	;;
+	cmp.eq p10,p8=r0,tmp
+	mov len1=len		// copy because of rotation
+(p8)	br.cond.dpnt .diff_align_copy_user
+	;;
+	// At this point we know we have more than 16 bytes to copy
+	// and also that both src and dest have the same alignment
+	// which may not be the one we want. So for now we must move
+	// forward slowly until we reach 16byte alignment: no need to
+	// worry about reaching the end of buffer.
+	//
+	EX(.failure_in1,(p6) ld1 val1[0]=[src1],1)	// 1-byte aligned
+(p6)	adds len1=-1,len1;;
+	tbit.nz p7,p0=src1,1
+	;;
+	EX(.failure_in1,(p7) ld2 val1[1]=[src1],2)	// 2-byte aligned
+(p7)	adds len1=-2,len1;;
+	tbit.nz p8,p0=src1,2
+	;;
+	//
+	// Stop bit not required after ld4 because if we fail on ld4
+	// we have never executed the ld1, therefore st1 is not executed.
+	//
+	EX(.failure_in1,(p8) ld4 val2[0]=[src1],4)	// 4-byte aligned
+	;;
+	EX(.failure_out,(p6) st1 [dst1]=val1[0],1)
+	tbit.nz p9,p0=src1,3
+	;;
+	//
+	// Stop bit not required after ld8 because if we fail on ld8
+	// we have never executed the ld2, therefore st2 is not executed.
+	//
+	EX(.failure_in1,(p9) ld8 val2[1]=[src1],8)	// 8-byte aligned
+	EX(.failure_out,(p7) st2 [dst1]=val1[1],2)
+(p8)	adds len1=-4,len1
+	;;
+	EX(.failure_out, (p8) st4 [dst1]=val2[0],4)
+(p9)	adds len1=-8,len1;;
+	shr.u cnt=len1,4		// number of 128-bit (2x64bit) words
+	;;
+	EX(.failure_out, (p9) st8 [dst1]=val2[1],8)
+	tbit.nz p6,p0=len1,3
+	cmp.eq p7,p0=r0,cnt
+	adds tmp=-1,cnt			// br.ctop is repeat/until
+(p7)	br.cond.dpnt .dotail		// we have less than 16 bytes left
+	;;
+	adds src2=8,src1
+	adds dst2=8,dst1
+	mov ar.lc=tmp
+	;;
+	//
+	// 16bytes/iteration
+	//
+2:
+	EX(.failure_in3,(p16) ld8 val1[0]=[src1],16)
+(p16)	ld8 val2[0]=[src2],16
+
+	EX(.failure_out, (EPI)	st8 [dst1]=val1[PIPE_DEPTH-1],16)
+(EPI)	st8 [dst2]=val2[PIPE_DEPTH-1],16
+	br.ctop.dptk 2b
+	;;			// RAW on src1 when fall through from loop
+	//
+	// Tail correction based on len only
+	//
+	// No matter where we come from (loop or test) the src1 pointer
+	// is 16 byte aligned AND we have less than 16 bytes to copy.
+	//
+.dotail:
+	EX(.failure_in1,(p6) ld8 val1[0]=[src1],8)	// at least 8 bytes
+	tbit.nz p7,p0=len1,2
+	;;
+	EX(.failure_in1,(p7) ld4 val1[1]=[src1],4)	// at least 4 bytes
+	tbit.nz p8,p0=len1,1
+	;;
+	EX(.failure_in1,(p8) ld2 val2[0]=[src1],2)	// at least 2 bytes
+	tbit.nz p9,p0=len1,0
+	;;
+	EX(.failure_out, (p6) st8 [dst1]=val1[0],8)
+	;;
+	EX(.failure_in1,(p9) ld1 val2[1]=[src1])	// only 1 byte left
+	mov ar.lc=saved_lc
+	;;
+	EX(.failure_out,(p7) st4 [dst1]=val1[1],4)
+	mov pr=saved_pr,0xffffffffffff0000
+	;;
+	EX(.failure_out, (p8)	st2 [dst1]=val2[0],2)
+	mov ar.pfs=saved_pfs
+	;;
+	EX(.failure_out, (p9)	st1 [dst1]=val2[1])
+	br.ret.sptk.many rp
+
+
+	//
+	// Here we handle the case where the byte by byte copy fails
+	// on the load.
+	// Several factors make the zeroing of the rest of the buffer kind of
+	// tricky:
+	//	- the pipeline: loads/stores are not in sync (pipeline)
+	//
+	//	  In the same loop iteration, the dst1 pointer does not directly
+	//	  reflect where the faulty load was.
+	//
+	//	- pipeline effect
+	//	  When you get a fault on load, you may have valid data from
+	//	  previous loads not yet store in transit. Such data must be
+	//	  store normally before moving onto zeroing the rest.
+	//
+	//	- single/multi dispersal independence.
+	//
+	// solution:
+	//	- we don't disrupt the pipeline, i.e. data in transit in
+	//	  the software pipeline will be eventually move to memory.
+	//	  We simply replace the load with a simple mov and keep the
+	//	  pipeline going. We can't really do this inline because
+	//	  p16 is always reset to 1 when lc > 0.
+	//
+.failure_in_pipe1:
+	sub ret0=endsrc,src1	// number of bytes to zero, i.e. not copied
+1:
+(p16)	mov val1[0]=r0
+(EPI)	st1 [dst1]=val1[PIPE_DEPTH-1],1
+	br.ctop.dptk 1b
+	;;
+	mov pr=saved_pr,0xffffffffffff0000
+	mov ar.lc=saved_lc
+	mov ar.pfs=saved_pfs
+	br.ret.sptk.many rp
+
+	//
+	// This is the case where the byte by byte copy fails on the load
+	// when we copy the head. We need to finish the pipeline and copy
+	// zeros for the rest of the destination. Since this happens
+	// at the top we still need to fill the body and tail.
+.failure_in_pipe2:
+	sub ret0=endsrc,src1	// number of bytes to zero, i.e. not copied
+2:
+(p16)	mov val1[0]=r0
+(EPI)	st1 [dst1]=val1[PIPE_DEPTH-1],1
+	br.ctop.dptk 2b
+	;;
+	sub len=enddst,dst1,1		// precompute len
+	br.cond.dptk.many .failure_in1bis
+	;;
+
+	//
+	// Here we handle the head & tail part when we check for alignment.
+	// The following code handles only the load failures. The
+	// main diffculty comes from the fact that loads/stores are
+	// scheduled. So when you fail on a load, the stores corresponding
+	// to previous successful loads must be executed.
+	//
+	// However some simplifications are possible given the way
+	// things work.
+	//
+	// 1) HEAD
+	// Theory of operation:
+	//
+	//  Page A   | Page B
+	//  ---------|-----
+	//          1|8 x
+	//	  1 2|8 x
+	//	    4|8 x
+	//	  1 4|8 x
+	//        2 4|8 x
+	//      1 2 4|8 x
+	//	     |1
+	//	     |2 x
+	//	     |4 x
+	//
+	// page_size >= 4k (2^12).  (x means 4, 2, 1)
+	// Here we suppose Page A exists and Page B does not.
+	//
+	// As we move towards eight byte alignment we may encounter faults.
+	// The numbers on each page show the size of the load (current alignment).
+	//
+	// Key point:
+	//	- if you fail on 1, 2, 4 then you have never executed any smaller
+	//	  size loads, e.g. failing ld4 means no ld1 nor ld2 executed
+	//	  before.
+	//
+	// This allows us to simplify the cleanup code, because basically you
+	// only have to worry about "pending" stores in the case of a failing
+	// ld8(). Given the way the code is written today, this means only
+	// worry about st2, st4. There we can use the information encapsulated
+	// into the predicates.
+	//
+	// Other key point:
+	//	- if you fail on the ld8 in the head, it means you went straight
+	//	  to it, i.e. 8byte alignment within an unexisting page.
+	// Again this comes from the fact that if you crossed just for the ld8 then
+	// you are 8byte aligned but also 16byte align, therefore you would
+	// either go for the 16byte copy loop OR the ld8 in the tail part.
+	// The combination ld1, ld2, ld4, ld8 where you fail on ld8 is impossible
+	// because it would mean you had 15bytes to copy in which case you
+	// would have defaulted to the byte by byte copy.
+	//
+	//
+	// 2) TAIL
+	// Here we now we have less than 16 bytes AND we are either 8 or 16 byte
+	// aligned.
+	//
+	// Key point:
+	// This means that we either:
+	//		- are right on a page boundary
+	//	OR
+	//		- are at more than 16 bytes from a page boundary with
+	//		  at most 15 bytes to copy: no chance of crossing.
+	//
+	// This allows us to assume that if we fail on a load we haven't possibly
+	// executed any of the previous (tail) ones, so we don't need to do
+	// any stores. For instance, if we fail on ld2, this means we had
+	// 2 or 3 bytes left to copy and we did not execute the ld8 nor ld4.
+	//
+	// This means that we are in a situation similar the a fault in the
+	// head part. That's nice!
+	//
+.failure_in1:
+	sub ret0=endsrc,src1	// number of bytes to zero, i.e. not copied
+	sub len=endsrc,src1,1
+	//
+	// we know that ret0 can never be zero at this point
+	// because we failed why trying to do a load, i.e. there is still
+	// some work to do.
+	// The failure_in1bis and length problem is taken care of at the
+	// calling side.
+	//
+	;;
+.failure_in1bis:		// from (.failure_in3)
+	mov ar.lc=len		// Continue with a stupid byte store.
+	;;
+5:
+	st1 [dst1]=r0,1
+	br.cloop.dptk 5b
+	;;
+	mov pr=saved_pr,0xffffffffffff0000
+	mov ar.lc=saved_lc
+	mov ar.pfs=saved_pfs
+	br.ret.sptk.many rp
+
+	//
+	// Here we simply restart the loop but instead
+	// of doing loads we fill the pipeline with zeroes
+	// We can't simply store r0 because we may have valid
+	// data in transit in the pipeline.
+	// ar.lc and ar.ec are setup correctly at this point
+	//
+	// we MUST use src1/endsrc here and not dst1/enddst because
+	// of the pipeline effect.
+	//
+.failure_in3:
+	sub ret0=endsrc,src1	// number of bytes to zero, i.e. not copied
+	;;
+2:
+(p16)	mov val1[0]=r0
+(p16)	mov val2[0]=r0
+(EPI)	st8 [dst1]=val1[PIPE_DEPTH-1],16
+(EPI)	st8 [dst2]=val2[PIPE_DEPTH-1],16
+	br.ctop.dptk 2b
+	;;
+	cmp.ne p6,p0=dst1,enddst	// Do we need to finish the tail ?
+	sub len=enddst,dst1,1		// precompute len
+(p6)	br.cond.dptk .failure_in1bis
+	;;
+	mov pr=saved_pr,0xffffffffffff0000
+	mov ar.lc=saved_lc
+	mov ar.pfs=saved_pfs
+	br.ret.sptk.many rp
+
+.failure_in2:
+	sub ret0=endsrc,src1
+	cmp.ne p6,p0=dst1,enddst	// Do we need to finish the tail ?
+	sub len=enddst,dst1,1		// precompute len
+(p6)	br.cond.dptk .failure_in1bis
+	;;
+	mov pr=saved_pr,0xffffffffffff0000
+	mov ar.lc=saved_lc
+	mov ar.pfs=saved_pfs
+	br.ret.sptk.many rp
+
+	//
+	// handling of failures on stores: that's the easy part
+	//
+.failure_out:
+	sub ret0=enddst,dst1
+	mov pr=saved_pr,0xffffffffffff0000
+	mov ar.lc=saved_lc
+
+	mov ar.pfs=saved_pfs
+	br.ret.sptk.many rp
+END(__copy_user)
+EXPORT_SYMBOL(__copy_user)