Re: [PATCH 15/44] kdbus: Simplify bitwise expression in kdbus_meta_get_mask()

Hi David, On Thu, Oct 08, 2015 at 04:24:30PM +0200, David Herrmann wrote: > Hi > > On Thu, Oct 8, 2015 at 1:31 PM, Sergei Zviagintsev wrote: > > Replace the expression with more concise and readable equivalent. It can > > be proven by opening parentheses: > > > > r & ~((p | i) & r) == r &

Re: [PATCH 15/44] kdbus: Simplify bitwise expression in kdbus_meta_get_mask()

Hi David, On Thu, Oct 08, 2015 at 04:24:30PM +0200, David Herrmann wrote: > Hi > > On Thu, Oct 8, 2015 at 1:31 PM, Sergei Zviagintsev wrote: > > Replace the expression with more concise and readable equivalent. It can > > be proven by opening parentheses: > > > > r & ~((p |

Re: [PATCH 15/44] kdbus: Simplify bitwise expression in kdbus_meta_get_mask()

Hi On Thu, Oct 8, 2015 at 1:31 PM, Sergei Zviagintsev 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

[PATCH 15/44] kdbus: Simplify bitwise expression in kdbus_meta_get_mask()

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 Signed-off-by: Sergei Zviagintsev --- ipc/kdbus/metadata.c | 2 +- 1 file

[PATCH 15/44] kdbus: Simplify bitwise expression in kdbus_meta_get_mask()

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 Signed-off-by: Sergei Zviagintsev ---

Re: [PATCH 15/44] kdbus: Simplify bitwise expression in kdbus_meta_get_mask()

Hi On Thu, Oct 8, 2015 at 1:31 PM, Sergei Zviagintsev 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) ==