Messages in this thread | | | Date | Tue, 5 Oct 2021 11:58:17 -0400 | From | Steven Rostedt <> | 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
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