Re: [RFC] bloody mess with __attribute__() syntax

[Al Viro <viro@ftp.linux.org.uk>]

On Thu, Jul 05, 2007 at 11:50:56AM -0700, Josh Triplett wrote:
> On Thu, 2007-07-05 at 17:43 +0100, Al Viro wrote:
> > On Thu, Jul 05, 2007 at 08:36:35AM -0700, Josh Triplett wrote:
> > > Wow.  Insane.  So these all declare the same type:
> > > __attribute__((foo)) T *v;
> > > T __attribute__((foo)) *v;
> > > T *__attribute__((foo)) v;
> > > ?  Specifically, they point to a foo-T, for convenient shooting?
> > 
> > They all give you foo-pointer-to-T.  
> > 	T (__attribute__((foo)) *v);
> > would give pointer-to-foo-T.
> 
> Doesn't that do exactly what we want, then?  If we say
> T __attribute__((noderef)) *v;
> , we want a noderef-pointer-to-T, not a pointer-to-noderef-T.  noderef
> should modify a pointer.

No.  int __user *v is pointer to noderef,address_space(1) int.  Same
as int const *v is pointer to const int.  Noderef is a property of
object being pointed to, _not_ the pointer itself.

And yes, I know that we store it ->modifiers of SYM_PTR - that saves us
a SYM_NODE we'd have to insert otherwise.  Same as with the rest of
qualifiers.

The same goes for address_space.  The same goes for const and volatile.

If you have struct foo {int x;}; struct foo __user *p; then &p->x will
be &((*p).x), i.e. &(<__user struct foo>.x), i.e. &(<__user int>), i.e.
int __user *.  __user is not a property of pointer; it couldn't work if
it would be.