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SubjectRe: [RFC][PATCH] rcu: Use typeof(p) instead of typeof(*p) *
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

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