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How does [=rcu_dereference_protected(ptr, 1)] differ from simple [=ptr]?

DISCUSSION Konstantin Andreev 1 participant 1 message May 15, 2019

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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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