copy_user.s

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	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 pr=saved_pr,0xffffffffffff0000
	mov ar.pfs=saved_pfs
	br.ret.dptk.few rp
	
	//
	// Beginning of long mempcy (i.e. > 16 bytes)
	//
long_copy_user:
	tbit.nz p6,p7=src1,0	// odd alignement
	and tmp=7,tmp
	;;
	cmp.eq p10,p8=r0,tmp
	mov len1=len		// copy because of rotation
(p8)	br.cond.dpnt.few 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.few .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.few 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.dptk.few 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.few 1b
	;;
	mov pr=saved_pr,0xffffffffffff0000
	mov ar.lc=saved_lc
	mov ar.pfs=saved_pfs
	br.ret.dptk.few 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.few 2b
	;;
	sub len=enddst,dst1,1		// precompute len
	br.cond.dptk.few 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 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=enddst,dst1	// number of bytes to zero, i.e. not copied
//	sub len=enddst,dst1,1
	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.few 5b	
	;;
skip_loop:
	mov pr=saved_pr,0xffffffffffff0000
	mov ar.lc=saved_lc
	mov ar.pfs=saved_pfs
	br.ret.dptk.few 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.few 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.few failure_in1bis	
	;;
	mov pr=saved_pr,0xffffffffffff0000
	mov ar.lc=saved_lc
	mov ar.pfs=saved_pfs
	br.ret.dptk.few rp

failure_in2:
	sub ret0=endsrc,src1	// number of bytes to zero, i.e. not copied
	;;
3:
(p16)	mov val1[0]=r0
(EPI)	st8 [dst1]=val1[PIPE_DEPTH-1],8
	br.ctop.dptk.few 3b
	;;
	cmp.ne p6,p0=dst1,enddst	// Do we need to finish the tail ?
	sub len=enddst,dst1,1		// precompute len
(p6)	br.cond.dptk.few failure_in1bis	
	;;
	mov pr=saved_pr,0xffffffffffff0000
	mov ar.lc=saved_lc
	mov ar.pfs=saved_pfs
	br.ret.dptk.few 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.dptk.few rp
END(__copy_user)