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>
This commit is contained in:
John David Anglin 2014-12-14 10:49:11 -05:00 committed by Helge Deller
parent b2776bf714
commit 45db07382a
1 changed files with 10 additions and 3 deletions

View File

@ -33,11 +33,18 @@
#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) ({ \
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; \
})