From 2b05d2b3b4d1e59e8710ec9274684d0d13eee34d Mon Sep 17 00:00:00 2001 From: Jesper Nilsson Date: Wed, 14 Nov 2007 17:01:30 -0800 Subject: [PATCH] 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: Jesper Nilsson Cc: Mikael Starvik Signed-off-by: Andrew Morton Signed-off-by: Linus Torvalds --- arch/cris/arch-v10/lib/usercopy.c | 314 +++++++++++++++++++------------------- 1 file changed, 157 insertions(+), 157 deletions(-) diff --git a/arch/cris/arch-v10/lib/usercopy.c b/arch/cris/arch-v10/lib/usercopy.c index a12c708..b8e6c04 100644 --- a/arch/cris/arch-v10/lib/usercopy.c +++ b/arch/cris/arch-v10/lib/usercopy.c @@ -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) -- 2.7.4