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From: Konstantin Andreev <andreev@swemel.ru>
To: kernelnewbies@kernelnewbies.org
Subject: How does [=rcu_dereference_protected(ptr, 1)] differ from simple [=ptr]?
Date: Wed, 15 May 2019 20:17:30 +0300	[thread overview]
Message-ID: <5CDC49AA.6090201@swemel.ru> (raw)

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.

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                 reply	other threads:[~2019-05-15 19:57 UTC|newest]

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