* How does [=rcu_dereference_protected(ptr, 1)] differ from simple [=ptr]?
@ 2019-05-15 17:17 Konstantin Andreev
0 siblings, 0 replies; only message in thread
From: Konstantin Andreev @ 2019-05-15 17:17 UTC (permalink / raw)
To: kernelnewbies
Hi,
according to the documentation, [rcu_dereference_protected( ptr, condition )] is the "update-side primitive" that reads the ptr, "allowing the code to verify that the required locks really are held".
Given that, I expect that literal `1' as `condition' would imply just "reading the ptr".
Indeed, the macro expansion of
= rcu_dereference_protected( ptr, 1 )
is roughly equals to just
= ({ (typeof(*ptr) *)ptr; })
i.e. simple
= ptr
that agrees with my expectation. So, coding [rcu_dereference_protected( ..., 1 )] looks like a waste of key presses.
However, expression [rcu_dereference_protected( ..., 1 )] routinely appears in the kernel sources, many times.
Why? Did I miss something?
Regards, Konstantin.
_______________________________________________
Kernelnewbies mailing list
Kernelnewbies@kernelnewbies.org
https://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
^ permalink raw reply [flat|nested] only message in thread
only message in thread, other threads:[~2019-05-15 19:57 UTC | newest]
Thread overview: (only message) (download: mbox.gz / follow: Atom feed)
-- links below jump to the message on this page --
2019-05-15 17:17 How does [=rcu_dereference_protected(ptr, 1)] differ from simple [=ptr]? Konstantin Andreev
This is a public inbox, see mirroring instructions
for how to clone and mirror all data and code used for this inbox;
as well as URLs for NNTP newsgroup(s).