On Thu, 2007-07-05 at 20:13 +0100, Al Viro wrote:
> 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.
OK, that seems inconsistent with what you said before. You said that
T __attribute__((foo)) *v;
gives you a foo-pointer-to-T. So shouldn't
int __attribute__((noderef)) *v;
give you a noderef-pointer-to-int?
> 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.
OK, that makes sense; address_space describes the actual storage of the
thing pointed to, not the pointer. It *could* describe the pointer, if
you had a pointer that resided in user address space, but that occurs
less often, and would use a different syntax.
However, noderef seems like a property of a pointer, hence why I
proposed the example I did. A warning should occur when you do
*(<noderef T *>v) to get a T, not when you do *(<* noderef T>v) to get a
noderef T.
- Josh Triplett
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