Hi

On Thu, Oct 8, 2015 at 1:31 PM, Sergei Zviagintsev <ser...@s15v.net> wrote:
> Replace the expression with more concise and readable equivalent. It can
> be proven by opening parentheses:
>
>     r & ~((p | i) & r) == r & (~(p | i) | ~r) ==
>         (r & ~(p | i)) | (r & ~r) == r & ~(p | i) == r & ~p & ~i

But why? The current code follows the description, and does exactly
the same. It shows that it merges the "provided" and "implied" masks,
and then extracts the flags that are missing compared to the required
mask.

I cannot follow why your code is more obvious?
David

> Signed-off-by: Sergei Zviagintsev <ser...@s15v.net>
> ---
>  ipc/kdbus/metadata.c | 2 +-
>  1 file changed, 1 insertion(+), 1 deletion(-)
>
> diff --git a/ipc/kdbus/metadata.c b/ipc/kdbus/metadata.c
> index 788b4d9c7ecb..61215a078359 100644
> --- a/ipc/kdbus/metadata.c
> +++ b/ipc/kdbus/metadata.c
> @@ -1321,7 +1321,7 @@ static u64 kdbus_meta_get_mask(struct pid *prv_pid, u64 
> prv_mask,
>          * the sender, but still requested by the receiver. If any are left,
>          * perform rather expensive /proc access checks for them.
>          */
> -       missing = req_mask & ~((prv_mask | impl_mask) & req_mask);
> +       missing = req_mask & ~prv_mask & ~impl_mask;
>         if (missing)
>                 proc_mask = kdbus_meta_proc_mask(prv_pid, req_pid, req_cred,
>                                                  missing);
> --
> 1.8.3.1
>
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