Thanks for response, but I did not understand your
point. Please read inline.
--- Ralf Baechle <email@example.com> wrote:
> On Sat, Feb 14, 2004 at 07:11:52AM -0800, Indigodfw
> > 2. Result of (C expression) should go into %xyz
> > register
> > So v->counter goes into %1, IOW ll from an int!
> > Does not make sense to me.
> > Why does it work, What am I missing?
> > I mean in general what is the expression for a m
> > constraint ptr (because I want ptr to be in
> > or *ptr (because I wanna tell compiler that *ptr
> > what gets changed)
> "m" gives you *something* suitable to address a
> memory object; that isn't
> necessarily a memory address. On MIPS it can't even
> be just an address
> in a register because "m" constraints are used with
> loads and stores and
> those only accept the offset(reg) addressing mode.
> If you want an address
> use something like "r" (&v->counter), then lw
Well, is not that what we want?
That is, we want to load (using ll) from &v->counter.
Should not the code have been
ll %0, 0(%1)
where %1 is "=m" (&v->counter)
The way I interpret it is:
a) %1 will contain address of v->counter
b) We would do ll (load with reservation/lock) from %1
which is &v->counter (and not from v->counter)
c)We want to tell the compiler that &v->counter is
output constraint and it may be modified. (Since
compiler does not look inside asm).
But I fear that with the syntax "=m" (&v->counter) we
are informing the compiler that this ptr itself may be
modified instead of its contents.
Do you Yahoo!?
Yahoo! Mail SpamGuard - Read only the mail you want.