This was my first try at #6, but I couldn't get it to work in z390. )-:
I don't like the multiply, though. I would have liked to just shift the bits
one to the left to do the multiply by 2, but I didn't figure out how to do that
yet.
START LH R1,=X'004A' R1 contains byte to convert
LA R2,7 R2 contains offset into BINNUM disp
XR R4,R4 R4 zero (MR overflow)
LA R5,1 R5 contains AND mask (counting)
NEXT1 LR R3,R5 R3 temporary register
NR R3,R1 Test if bit is on
BZ ZERO ..No, write "0"
L R3,=X'000000F1' Move "1" to R3
STC R3,BINNUM(R2) Move to BINNUM
B SKIP
ZERO L R3,=X'000000F0' Move "0" to R3
STC R3,BINNUM(R2) Move to BINNUM
SKIP LA R3,2 Set R3 to 2 for multiply
MR R4,R3 Double R4 to shift bit 1 to left
LTR R2,R2 Is it zero?
BZ PRINTIT Yes, done, print results
BCTR R2,R0 Decrement R2
B NEXT1 Go get next one
*
PRINTIT TPUT BINNUM,L'BINNUM TPUT Results
*
*
BINNUM DC CL8'$$$$$$$$' Converted Number
-----Original Message-----
From: IBM Mainframe Discussion List on behalf of Binyamin Dissen
Sent: Wed 12/12/2007 8:52 PM
To: [email protected]
Subject: Re: Mainframe Assembler Coding Contest
Constraints are not well defined.
If one can use an 8*256 byte table .... 3 silicon instructions (4, if LB is in
silicon) - not tested
...entry logic
LB R1,BYTE
SLL R1,3
LA R2,TABLE+8*128(R1)
MVC RESULT,0(R2)
...exit logic
TABLE DC C'10000000'
DC C'10000001'
...
DC C'11111111'
DC C'00000000'
DC C'00000001'
...
DC C'01111111'
--
Binyamin Dissen <[EMAIL PROTECTED]>
http://www.dissensoftware.com
Director, Dissen Software, Bar & Grill - Israel
Should you use the mailblocks package and expect a response from me,
you should preauthorize the dissensoftware.com domain.
I very rarely bother responding to challenge/response systems,
especially those from irresponsible companies.
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