RE: Bit Operations

No, that's the job of the AND operation.

You're getting very close.  X'10' does indeed equal B'0001 0000'.  In
hexadecimal, each hex digit represents four binary digits (in fact,
that's what bit means: Binary digIT).

Okay, now you need to take a look at the AND operation.  There are two
operands and the result, right?  Okay, the first operand is the original
value, and the second operand is the "mask".  Basically, the AND will
turn off every bit in the original value if the corresponding bit in the
mask is zero.  So:

B'0111 0000' AND B'0001 0000' = B'0001 0000'

Another way to look at this is:

B'0111 0000'
B'0001 0000'
------------
B'0001 0000'

See how bits one and two get turned off, but bit three stays on?  That's
because with an AND operation, in order for the bit to be on in the
result, the bit must be on in both the original value AND the mask (thus
the name of the operation).

So, let's look at a couple more examples:

B'0001 0000' AND B'0001 0000' = B'0001 0000'
B'0001 1111' AND B'0001 0000' = B'0001 0000'
B'0101 0101' AND B'0001 0000' = B'0001 0000'
B'1111 1111' AND B'0001 0000' = B'0001 0000'

Get the idea?  Basically, the AND strips out anything except where you
have bits on in the mask.  So, in this particular case, as long as bit
one is on, the result will be B'0001 0000'.  Similarly, anything with
bit one off will end up as B'0000 0000':

B'0000 0000' AND B'0001 0000' = B'0000 0000'
B'0000 1111' AND B'0001 0000' = B'0000 0000'
B'0000 0101' AND B'0001 0000' = B'0000 0000'
B'1110 1111' AND B'0001 0000' = B'0000 0000'

Because there is only one bit on in your mask, there are only two
possible results: if bit three is on, the result is B'0001 0000', if bit
three is off, the result is B'0000 0000'.  So, the way to check if bit
three is on is to AND it with a mask of B'0001 0000' and then see if the
result is also B'0001 0000'.  Since B'0001 0000' is equivalent to X'10',
you get the following:

bit3ison = (%BitAnd(YourField:X'10') = X'10');

In general, to test that one or more bits are all on, create a mask with
those bits on.  Then, AND the original value with the mask and make sure
the result is the same as the mask.  (If ALL the bits are off, the
result will be zero; if ANY of the bits are on the result will be
non-zero.)

Finally, typically, I'd use a named constant for this.  For example:

d BIT_THREE       c                   x'10'
d BITS_0123       c                   x'F0'

Then I could say:

bit3ison = (%BitAnd(YourField:BIT_THREE) = BIT_THREE);
nohighbits = (%BitAnd(YourField:BITS_0123) = 0);

Hope this helps!

Joe


> From: Fisher, Don
> 
> A little.  I guess I'm a little slow when it comes to this stuff.  I
don't
> understand why X'10' is being used both as the comparison value and
the
> identification of the bit, unless X'10' is the hexadecimal equivalent
of
> '0001 0000'.
> 
> If that's the case, I either need to figure out all the possible
> hexadecimal
> values where the third bit (bit number 2) can be on or a method of
> determining the status of that third bit directly.

> Sent: Tuesday, May 31, 2005 7:17 AM
> To: 'RPG programming on the AS400 / iSeries'
> Subject: RE: Bit Operations