Re: In line ASM magic? What is this?

From: George Anzinger (george@mvista.com)
Date: Wed Nov 15 2000 - 17:32:07 EST


Timur Tabi wrote:
>
> ** Reply to message from George Anzinger <george@mvista.com> on Wed, 15 Nov
> 2000 12:55:46 -0800
>
> > I am trying to understand what is going on in the following code. The
> > reference for %2, i.e. "m"(*__xg(ptr)) seems like magic (from
> > .../include/i386/system.h). At the same time, the code "m" (*mem) from
> > the second __asm__ below (my code) seems to generate the required asm
> > code. Before I go with the simple version, could someone tell me why?
> > Inquiring minds want to know.
> >
> > struct __xchg_dummy { unsigned long a[100]; };
> > #define __xg(x) ((struct __xchg_dummy *)(x))
> >
> > __asm__ __volatile__(LOCK_PREFIX "cmpxchgl %b1,%2"
> > : "=a"(prev)
> > : "q"(new), "m"(*__xg(ptr)), "0"(old)
> > : "memory");
> >
> >
> > __asm__ __volatile__(
> > LOCK "cmpxchgl %1,%2\n\t"
> > :"=a" (result)
> > :"r" (new),
> > "m" (*mem),
> > "a0" (test)
> > : "memory");
>
> I've been a lot of gcc inline asm recently, and I still consider it a black
> art. There are times when I just throw in what I think makes sense, and then
> look at the code the compiler generated. If it's wrong, I try something else.
>
> Both versions look correct to me. The "m" simply tells the compiler that
> __xg(ptr) is a memory location, and the contents of that memory location should
> NOT be copied to a register. The confusion occurs because its unintuitive that
> the "*" is required. Otherwise, it would have been "r", which basically tells
> the compiler to copy the contents to a register first.
>
I know the feeling. I am currently strugling with "inconsistant
constraints". Still, I must assume that form 1 was used instead of 2
for some reason....

George
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