Vince Vielhaber wrote:
> Do we have any kind of bitwise AND?
Non directly.
> select foo from bar where (foo AND 2);
Ok, you _can_ do this; it's just a big pain to do so.
First, decompose any bitwise AND into TESTBIT and logical AND
operators. TESTBIT is the same as AND, just having a single bit set.
In your example, AND 2 is already a single bit. But, I'm going to
illustrate the general solution: suppose we have AND 7 -- decompose into
(foo TESTBIT 4) AND (foo TESTBIT 2) AND (foo TESTBIT 1).
Now, rewrite that to:
((foo % 8)/4)*((foo % 4)/2)*((foo % 2)/1) -- if the result is greater
than zero, the logical AND is true.
For bitwise OR, substitute integer + for integer * above.
For your case, write the query:
select foo from bar where ((foo % 4)/2)>0
AFAIK and have tested, that should work the way you think it should. (I
knew those exercise in Z80 machine language would come in handy! :-))
--
Lamar Owen
WGCR Internet Radio
1 Peter 4:11