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CRISv10 memset library add lineendings to asm

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Add \n\ at end of lines inside asm statement to avoid warning.

No change except adding \n\ to end of line and correcting
whitespace has been done.
Removes warning about multi-line string literals when compiling
arch/cris/arch-v10/lib/memset.c

Signed-off-by: Jesper Nilsson <jesper.nilsson@axis.com>
Cc: Mikael Starvik <mikael.starvik@axis.com>
Signed-off-by: Andrew Morton <akpm@linux-foundation.org>
Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
This commit is contained in:
Jesper Nilsson 2007-11-14 17:01:32 -08:00 committed by Linus Torvalds
parent 341ac6e4be
commit 2e2cd8bad6
1 changed files with 41 additions and 41 deletions
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82
arch/cris/arch-v10/lib/memset.c
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@ -66,7 +66,7 @@ void *memset(void *pdst,
{
register char *dst __asm__ ("r13") = pdst;
/* This is NONPORTABLE, but since this whole routine is */
/* grossly nonportable that doesn't matter. */
@ -110,52 +110,52 @@ void *memset(void *pdst,
If you want to check that the allocation was right; then
check the equalities in the first comment. It should say
"r13=r13, r12=r12, r11=r11" */
__asm__ volatile ("
;; Check that the following is true (same register names on
;; both sides of equal sign, as in r8=r8):
;; %0=r13, %1=r12, %4=r11
;;
;; 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]
move.d $r11,$r0
move.d $r11,$r1
move.d $r11,$r2
move.d $r11,$r3
move.d $r11,$r4
move.d $r11,$r5
move.d $r11,$r6
move.d $r11,$r7
move.d $r11,$r8
move.d $r11,$r9
move.d $r11,$r10
;; 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+]
addq 12*4,$r12 ;; compensate for last loop underflowing n
;; Restore registers from stack
movem [$sp+],$r10"
__asm__ volatile ("\n\
;; Check that the following is true (same register names on \n\
;; both sides of equal sign, as in r8=r8): \n\
;; %0=r13, %1=r12, %4=r11 \n\
;; \n\
;; Save the registers we'll clobber in the movem process \n\
;; on the stack. Don't mention them to gcc, it will only be \n\
;; upset. \n\
subq 11*4,$sp \n\
movem $r10,[$sp] \n\
\n\
move.d $r11,$r0 \n\
move.d $r11,$r1 \n\
move.d $r11,$r2 \n\
move.d $r11,$r3 \n\
move.d $r11,$r4 \n\
move.d $r11,$r5 \n\
move.d $r11,$r6 \n\
move.d $r11,$r7 \n\
move.d $r11,$r8 \n\
move.d $r11,$r9 \n\
move.d $r11,$r10 \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\
\n\
addq 12*4,$r12 ;; compensate for last loop underflowing n \n\
\n\
;; Restore registers from stack \n\
movem [$sp+],$r10"
/* Outputs */ : "=r" (dst), "=r" (n)
/* Inputs */ : "0" (dst), "1" (n), "r" (lc));
}
/* Either we directly starts copying, using dword copying
in a loop, or we copy as much as possible with 'movem'
in a loop, or we copy as much as possible with 'movem'
and then the last block (<44 bytes) is copied here.
This will work since 'movem' will have updated src,dst,n. */