Commit 2b05d2b3 authored by Jesper Nilsson's avatar Jesper Nilsson Committed by Linus Torvalds
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CRISv10 usercopy library add lineendings to asm



Removes warning when compiling arch/cris/arch-v10/lib/usercopy.c

No change except adding \n\ on the end of the lines has been done.
Removes warning about multi-line string literals.

Signed-off-by: default avatarJesper Nilsson <jesper.nilsson@axis.com>
Cc: Mikael Starvik <mikael.starvik@axis.com>
Signed-off-by: default avatarAndrew Morton <akpm@linux-foundation.org>
Signed-off-by: default avatarLinus Torvalds <torvalds@linux-foundation.org>
parent 3eed6393
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+157 −157
Original line number Diff line number Diff line
@@ -92,58 +92,58 @@ __copy_user (void __user *pdst, const void *psrc, unsigned long pn)
	.ifnc %0%1%2%3,$r13$r11$r12$r10					\n\
	.err								\n\
	.endif								\n\

	;; Save the registers we'll use in the movem process
	;; on the stack.
	subq	11*4,$sp
	movem	$r10,[$sp]

	;; Now we've got this:
	;; r11 - src
	;; r13 - dst
	;; r12 - n

	;; Update n for the first loop
	subq	44,$r12

; Since the noted PC of a faulting instruction in a delay-slot of a taken
; branch, is that of the branch target, we actually point at the from-movem
; for this case.  There is no ambiguity here; if there was a fault in that
; instruction (meaning a kernel oops), the faulted PC would be the address
; after *that* movem.

0:
	movem	[$r11+],$r10
	subq   44,$r12
	bge	0b
	movem	$r10,[$r13+]
1:
	addq   44,$r12  ;; compensate for last loop underflowing n

	;; Restore registers from stack
	movem [$sp+],$r10
2:
	.section .fixup,\"ax\"

; To provide a correct count in r10 of bytes that failed to be copied,
; we jump back into the loop if the loop-branch was taken.  There is no
; performance penalty for sany use; the program will segfault soon enough.

3:
	move.d [$sp],$r10
	addq 44,$r10
	move.d $r10,[$sp]
	jump 0b
4:
	movem [$sp+],$r10
	addq 44,$r10
	addq 44,$r12
	jump 2b

	.previous
	.section __ex_table,\"a\"
	.dword 0b,3b
	.dword 1b,4b
									\n\
	;; Save the registers we'll use in the movem process		\n\
	;; on the stack.						\n\
	subq	11*4,$sp						\n\
	movem	$r10,[$sp]						\n\
									\n\
	;; Now we've got this:						\n\
	;; r11 - src							\n\
	;; r13 - dst							\n\
	;; r12 - n							\n\
									\n\
	;; Update n for the first loop					\n\
	subq	44,$r12							\n\
									\n\
; Since the noted PC of a faulting instruction in a delay-slot of a taken \n\
; branch, is that of the branch target, we actually point at the from-movem \n\
; for this case.  There is no ambiguity here; if there was a fault in that \n\
; instruction (meaning a kernel oops), the faulted PC would be the address \n\
; after *that* movem.							\n\
									\n\
0:									\n\
	movem	[$r11+],$r10						\n\
	subq   44,$r12							\n\
	bge	0b							\n\
	movem	$r10,[$r13+]						\n\
1:									\n\
	addq   44,$r12  ;; compensate for last loop underflowing n	\n\
									\n\
	;; Restore registers from stack					\n\
	movem [$sp+],$r10						\n\
2:									\n\
	.section .fixup,\"ax\"						\n\
									\n\
; To provide a correct count in r10 of bytes that failed to be copied,	\n\
; we jump back into the loop if the loop-branch was taken.  There is no	\n\
; performance penalty for sany use; the program will segfault soon enough.\n\
									\n\
3:									\n\
	move.d [$sp],$r10						\n\
	addq 44,$r10							\n\
	move.d $r10,[$sp]						\n\
	jump 0b								\n\
4:									\n\
	movem [$sp+],$r10						\n\
	addq 44,$r10							\n\
	addq 44,$r12							\n\
	jump 2b								\n\
									\n\
	.previous							\n\
	.section __ex_table,\"a\"					\n\
	.dword 0b,3b							\n\
	.dword 1b,4b							\n\
	.previous"

     /* Outputs */ : "=r" (dst), "=r" (src), "=r" (n), "=r" (retn)
@@ -253,59 +253,59 @@ __copy_user_zeroing (void __user *pdst, const void *psrc, unsigned long pn)
       If you want to check that the allocation was right; then
       check the equalities in the first comment.  It should say
       "r13=r13, r11=r11, r12=r12" */
    __asm__ volatile ("
    __asm__ volatile ("\n\
	.ifnc %0%1%2%3,$r13$r11$r12$r10					\n\
	.err								\n\
	.endif								\n\

	;; Save the registers we'll use in the movem process
	;; on the stack.
	subq	11*4,$sp
	movem	$r10,[$sp]

	;; Now we've got this:
	;; r11 - src
	;; r13 - dst
	;; r12 - n

	;; Update n for the first loop
	subq	44,$r12
0:
	movem	[$r11+],$r10
1:
	subq   44,$r12
	bge	0b
	movem	$r10,[$r13+]

	addq   44,$r12  ;; compensate for last loop underflowing n

	;; Restore registers from stack
	movem [$sp+],$r10
4:
	.section .fixup,\"ax\"

;; Do not jump back into the loop if we fail.  For some uses, we get a
;; page fault somewhere on the line.  Without checking for page limits,
;; we don't know where, but we need to copy accurately and keep an
;; accurate count; not just clear the whole line.  To do that, we fall
;; down in the code below, proceeding with smaller amounts.  It should
;; be kept in mind that we have to cater to code like what at one time
;; was in fs/super.c:
;;  i = size - copy_from_user((void *)page, data, size);
;; which would cause repeated faults while clearing the remainder of
;; the SIZE bytes at PAGE after the first fault.
;; A caveat here is that we must not fall through from a failing page
;; to a valid page.

3:
	movem  [$sp+],$r10
	addq	44,$r12 ;; Get back count before faulting point.
	subq	44,$r11 ;; Get back pointer to faulting movem-line.
	jump	4b	;; Fall through, pretending the fault didn't happen.

	.previous
	.section __ex_table,\"a\"
	.dword 1b,3b
									\n\
	;; Save the registers we'll use in the movem process		\n\
	;; on the stack.						\n\
	subq	11*4,$sp						\n\
	movem	$r10,[$sp]						\n\
									\n\
	;; Now we've got this:						\n\
	;; r11 - src							\n\
	;; r13 - dst							\n\
	;; r12 - n							\n\
									\n\
	;; Update n for the first loop					\n\
	subq	44,$r12							\n\
0:									\n\
	movem	[$r11+],$r10						\n\
1:									\n\
	subq   44,$r12							\n\
	bge	0b							\n\
	movem	$r10,[$r13+]						\n\
									\n\
	addq   44,$r12  ;; compensate for last loop underflowing n	\n\
									\n\
	;; Restore registers from stack					\n\
	movem [$sp+],$r10						\n\
4:									\n\
	.section .fixup,\"ax\"						\n\
									\n\
;; Do not jump back into the loop if we fail.  For some uses, we get a	\n\
;; page fault somewhere on the line.  Without checking for page limits,	\n\
;; we don't know where, but we need to copy accurately and keep an	\n\
;; accurate count; not just clear the whole line.  To do that, we fall	\n\
;; down in the code below, proceeding with smaller amounts.  It should	\n\
;; be kept in mind that we have to cater to code like what at one time	\n\
;; was in fs/super.c:							\n\
;;  i = size - copy_from_user((void *)page, data, size);		\n\
;; which would cause repeated faults while clearing the remainder of	\n\
;; the SIZE bytes at PAGE after the first fault.			\n\
;; A caveat here is that we must not fall through from a failing page	\n\
;; to a valid page.							\n\
									\n\
3:									\n\
	movem  [$sp+],$r10						\n\
	addq	44,$r12 ;; Get back count before faulting point.	\n\
	subq	44,$r11 ;; Get back pointer to faulting movem-line.	\n\
	jump	4b	;; Fall through, pretending the fault didn't happen.\n\
									\n\
	.previous							\n\
	.section __ex_table,\"a\"					\n\
	.dword 1b,3b							\n\
	.previous"

     /* Outputs */ : "=r" (dst), "=r" (src), "=r" (n), "=r" (retn)
@@ -425,64 +425,64 @@ __do_clear_user (void __user *pto, unsigned long pn)
      If you want to check that the allocation was right; then
      check the equalities in the first comment.  It should say
      something like "r13=r13, r11=r11, r12=r12". */
    __asm__ volatile ("
    __asm__ volatile ("\n\
	.ifnc %0%1%2,$r13$r12$r10					\n\
	.err								\n\
	.endif								\n\

	;; Save the registers we'll clobber in the movem process
	;; on the stack.  Don't mention them to gcc, it will only be
	;; upset.
	subq	11*4,$sp
	movem	$r10,[$sp]

	clear.d $r0
	clear.d $r1
	clear.d $r2
	clear.d $r3
	clear.d $r4
	clear.d $r5
	clear.d $r6
	clear.d $r7
	clear.d $r8
	clear.d $r9
	clear.d $r10
	clear.d $r11

	;; Now we've got this:
	;; r13 - dst
	;; r12 - n

	;; Update n for the first loop
	subq	12*4,$r12
0:
	subq   12*4,$r12
	bge	0b
	movem	$r11,[$r13+]
1:
	addq   12*4,$r12        ;; compensate for last loop underflowing n

	;; Restore registers from stack
	movem [$sp+],$r10
2:
	.section .fixup,\"ax\"
3:
	move.d [$sp],$r10
	addq 12*4,$r10
	move.d $r10,[$sp]
	clear.d $r10
	jump 0b

4:
	movem [$sp+],$r10
	addq 12*4,$r10
	addq 12*4,$r12
	jump 2b

	.previous
	.section __ex_table,\"a\"
	.dword 0b,3b
	.dword 1b,4b
									\n\
	;; Save the registers we'll clobber in the movem process	\n\
	;; on the stack.  Don't mention them to gcc, it will only be	\n\
	;; upset.							\n\
	subq	11*4,$sp						\n\
	movem	$r10,[$sp]						\n\
									\n\
	clear.d $r0							\n\
	clear.d $r1							\n\
	clear.d $r2							\n\
	clear.d $r3							\n\
	clear.d $r4							\n\
	clear.d $r5							\n\
	clear.d $r6							\n\
	clear.d $r7							\n\
	clear.d $r8							\n\
	clear.d $r9							\n\
	clear.d $r10							\n\
	clear.d $r11							\n\
									\n\
	;; Now we've got this:						\n\
	;; r13 - dst							\n\
	;; r12 - n							\n\
									\n\
	;; Update n for the first loop					\n\
	subq	12*4,$r12						\n\
0:									\n\
	subq   12*4,$r12						\n\
	bge	0b							\n\
	movem	$r11,[$r13+]						\n\
1:									\n\
	addq   12*4,$r12        ;; compensate for last loop underflowing n\n\
									\n\
	;; Restore registers from stack					\n\
	movem [$sp+],$r10						\n\
2:									\n\
	.section .fixup,\"ax\"						\n\
3:									\n\
	move.d [$sp],$r10						\n\
	addq 12*4,$r10							\n\
	move.d $r10,[$sp]						\n\
	clear.d $r10							\n\
	jump 0b								\n\
									\n\
4:									\n\
	movem [$sp+],$r10						\n\
	addq 12*4,$r10							\n\
	addq 12*4,$r12							\n\
	jump 2b								\n\
									\n\
	.previous							\n\
	.section __ex_table,\"a\"					\n\
	.dword 0b,3b							\n\
	.dword 1b,4b							\n\
	.previous"

     /* Outputs */ : "=r" (dst), "=r" (n), "=r" (retn)