ss: Drop silly assignment

An expression of the form '(a | b) & b' will evaluate to the value of b
for any value of a or b.

Signed-off-by: Phil Sutter <phil@nwl.cc>

diff --git a/misc/ss.c b/misc/ss.c
index 38cf3312a746e6d5d81fa2eee6453aa6a9291c41..d6090018c5dbbf16058dca19ed17627984d54072 100644 (file)
--- a/misc/ss.c
+++ b/misc/ss.c
@@ -267,7 +267,7 @@ static void filter_default_dbs(struct filter *f)
 static void filter_states_set(struct filter *f, int states)
 {
        if (states)
-               f->states = (f->states | states) & states;
+               f->states = states;
 }
 
 static void filter_merge_defaults(struct filter *f)