From mboxrd@z Thu Jan 1 00:00:00 1970 From: Al Viro Subject: Re: [RFC] bloody mess with __attribute__() syntax Date: Thu, 5 Jul 2007 20:13:59 +0100 Message-ID: <20070705191358.GQ21478@ftp.linux.org.uk> References: <20070705093528.GK21478@ftp.linux.org.uk> <468D1003.1050901@freedesktop.org> <20070705164334.GM21478@ftp.linux.org.uk> <1183661456.2604.43.camel@josh-work.beaverton.ibm.com> Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii Return-path: Received: from zeniv.linux.org.uk ([195.92.253.2]:41953 "EHLO ZenIV.linux.org.uk" rhost-flags-OK-OK-OK-OK) by vger.kernel.org with ESMTP id S1760676AbXGETOB (ORCPT ); Thu, 5 Jul 2007 15:14:01 -0400 Content-Disposition: inline In-Reply-To: <1183661456.2604.43.camel@josh-work.beaverton.ibm.com> Sender: linux-sparse-owner@vger.kernel.org List-Id: linux-sparse@vger.kernel.org To: Josh Triplett Cc: Josh Triplett , linux-sparse@vger.kernel.org, Linus Torvalds , linux-kernel@vger.kernel.org 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.