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parisc: fix out-of-register compiler error in ldcw inline assembler function

The __ldcw macro has a problem when its argument needs to be reloaded from
memory. The output memory operand and the input register operand both need to
be reloaded using a register in class R1_REGS when generating 64-bit code.
This fails because there's only a single register in the class. Instead, use a
memory clobber. This also makes the __ldcw macro a compiler memory barrier.

Signed-off-by: John David Anglin <dave.anglin@bell.net>
Cc: <stable@vger.kernel.org>        [3.13+]
Signed-off-by: Helge Deller <deller@gmx.de>
John David Anglin 10 years ago
parent
commit
45db07382a
1 changed files with 10 additions and 3 deletions
  1. 10 3
      arch/parisc/include/asm/ldcw.h

+ 10 - 3
arch/parisc/include/asm/ldcw.h

@@ -33,11 +33,18 @@
 
 
 #endif /*!CONFIG_PA20*/
 #endif /*!CONFIG_PA20*/
 
 
-/* LDCW, the only atomic read-write operation PA-RISC has. *sigh*.  */
+/* LDCW, the only atomic read-write operation PA-RISC has. *sigh*.
+   We don't explicitly expose that "*a" may be written as reload
+   fails to find a register in class R1_REGS when "a" needs to be
+   reloaded when generating 64-bit PIC code.  Instead, we clobber
+   memory to indicate to the compiler that the assembly code reads
+   or writes to items other than those listed in the input and output
+   operands.  This may pessimize the code somewhat but __ldcw is
+   usually used within code blocks surrounded by memory barriors.  */
 #define __ldcw(a) ({						\
 #define __ldcw(a) ({						\
 	unsigned __ret;						\
 	unsigned __ret;						\
-	__asm__ __volatile__(__LDCW " 0(%2),%0"			\
-		: "=r" (__ret), "+m" (*(a)) : "r" (a));		\
+	__asm__ __volatile__(__LDCW " 0(%1),%0"			\
+		: "=r" (__ret) : "r" (a) : "memory");		\
 	__ret;							\
 	__ret;							\
 })
 })