RE: Free-form Equivalent to BITOFF?

[Paul Jackson <pjackson@xxxxxxxxxx>]

Thank you Scott and Daniel!, 

I think I get the gist of it now.

-----Original Message-----
From: rpg400-l-bounces@xxxxxxxxxxxx
[mailto:rpg400-l-bounces@xxxxxxxxxxxx]On Behalf Of Scott Klement
Sent: Wednesday, February 16, 2005 3:03 PM
To: RPG programming on the AS400 / iSeries
Subject: Re: Free-form Equivalent to BITOFF?


Hi,

> I am trying to convert a statement that turns off the first four bits in a
> single byte field to free-form and am unsure of the best way to do that.
>
> Original statement:
> C                   BITOFF    '0123'        BYTE

Create a hex field containing the bits that you want to keep on. The 
%bitand() BIF will return a new byte where only those bits are copied from 
the original.

The old opcodes (like your example above) confuse me, so I hope I'm right, 
but this is what I suggest:

    byte = %bitand(x'0F': BYTE);

How this works is based on copying the bits that are on in BOTH bytes to 
the result. For the sake of example, let's say that BYTE contains the 
letter R, which is x'D9' in EBCDIC (I just picked this as a random value)
You'll have the following in binary:

    x'0F' = 00001111
    x'D9' = 11011001

So, it copies the bits that are in on BOTH fields, yielding:
            00001001

In other words, since I used x'0F' (which has the first 4 bits off, and 
the last 4 bits on) it copies just the last 4 bits of the field, and sets 
the rest of them off.

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