Re: set_bit() is broken on i386?

[Chuck Ebbert <76306.1226@compuserve.com>]

In-Reply-To: <20060120183857.188ef516.akpm@osdl.org>

On Fri, 20 Jan 2006, Andrew Morton wrote:

> We need to somehow tell the compiler "this assembly statement altered
> memory and you can't cache memory contents across it".  That's what
> "memory" (ie: barrier()) does.  I don't think there's a way of telling gcc
> _what_ memory was clobbered - just "all of memory".

I think you can do that by specifying an output operand that you
never use in your assembler code, e.g. changing this:

|       __asm__ __volatile__( "lock ; "
|               "btsl %1,%0"
|               :"=m" (ADDR)
|               :"Ir" (nr));

to this:

| #define LONGBITS (8 * sizeof(unsigned long))
|
|       __asm__ __volatile__( "lock ; "
|               "btsl %2,%1"
|               :"=m"(*(&ADDR + nr/LONGBITS))
|               :"m" (ADDR), "Ir" (nr));

fixes my example program by telling the compiler what memory location
is altered.  (Note that %0 is never used inside the asm code.)
So iff 'nr' is a constant you can clobber specific memory locations.
-- 
Chuck
Currently reading: _Sun Of Suns_ by Karl Schroeder
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