I've been testing all... Might as well come out of the dark...

You pass ! 

-----Original Message-----
From: IBM Mainframe Discussion List [mailto:[EMAIL PROTECTED] On
Behalf Of Art Celestini
Sent: Wednesday, December 12, 2007 8:52 PM
To: IBM-MAIN@BAMA.UA.EDU
Subject: Re: Mainframe Assembler Coding Contest

Well, just for the sport of it, I think this requires even less (not
tested):

         LHI   R1,8
         IC    R2,BYTE
STC      STC   R2,RESULT-1(R1)
         SRL   R2,1
         BCT   R1,STC
         NC    RESULT,=8X'01'
         OC    RESULT,=8X'F0'
*
BYTE     DC    B'01110101'        Sample input
RESULT   DS    CL8                Output

   32  Instructions
   16  Literals
    1  Input Area
    8  Output Area

   57  Total Bytes
    7  Total Instructions

BTW, John, your second instruction below refers to WORK8, which doesn't
seem to be defined.

-- Art


At 04:30 PM 12/12/2007, John P Baker wrote:
  
>I have not tested it, but the following code should be the most compact
form of generating a binary character string from a byte value (Problem
#6):
>
>         MVC   OBUFF,IBYTE
>         MVC   OBUFF+1(7),WORK8
>         NC    OBUFF,=X'8040201008040201'
>         MVO   WORK5,OBUFF(4)
>         MVC   OBUFF(4),WORK5
>         TR    OBUFF,=C'011111111'
>
>IBYTE    DC    B'01110101'         Sample input value
>WORK5    DC    X'0000000000'
>OBUFF    DS    CL8                 Output value
>
>Summary of storage requirements:
>
>  36  Instructions
>  17  Literals
>   1  Input Buffer
>   5  Work Area
>   8  Output Area
>
>  67  Total Bytes
>   6  Total Instructions



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Art Celestini       Celestini Development Services
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