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Date:   Tue, 5 Oct 2021 11:58:17 -0400
From:   Steven Rostedt <rostedt@goodmis.org>
To:     Mathieu Desnoyers <mathieu.desnoyers@efficios.com>
Cc:     linux-kernel <linux-kernel@vger.kernel.org>,
        Linus Torvalds <torvalds@linux-foundation.org>,
        Paul <paulmck@linux.vnet.ibm.com>,
        Josh Triplett <josh@joshtriplett.org>,
        Lai Jiangshan <jiangshanlai@gmail.com>,
        "Joel Fernandes, Google" <joel@joelfernandes.org>,
        Pablo Neira Ayuso <pablo@netfilter.org>,
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        netfilter-devel <netfilter-devel@vger.kernel.org>,
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Subject: Re: [RFC][PATCH] rcu: Use typeof(p) instead of typeof(*p) *
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On Tue, 5 Oct 2021 11:15:12 -0400 (EDT)
Mathieu Desnoyers <mathieu.desnoyers@efficios.com> wrote:

> ----- On Oct 5, 2021, at 9:47 AM, rostedt rostedt@goodmis.org wrote:
> [...]
> > #define rcu_dereference_raw(p) \
> > ({ \
> > 	/* Dependency order vs. p above. */ \
> > 	typeof(p) ________p1 = READ_ONCE(p); \
> > -	((typeof(*p) __force __kernel *)(________p1)); \
> > +	((typeof(p) __force __kernel)(________p1)); \
> > })  
> 
> AFAIU doing so removes validation that @p is indeed a pointer, so a user might mistakenly
> try to use rcu_dereference() on an integer, and get away with it. I'm not sure we want to
> loosen this check. I wonder if there might be another way to achieve the same check without
> requiring the structure to be declared, e.g. with __builtin_types_compatible_p ?

Is that really an issue? Because you would be assigning it to an integer.


	x = rcu_dereference_raw(y);

And that just makes 'x' a copy of 'y' and not really a reference to it, thus
if you don't have a pointer, it's just a fancy READ_ONCE(y).

-- Steve