Click to See Complete Forum and Search --> : Challenge: What is the result of this prog? (TOUGH)
bugfix
01-29-2002, 11:43 AM
R8 is address of the program in memory
Assume R6 is X'00000010' at the start
What does R6 equal at the end of this program?
18CF50E0C03459608024477080165860802847F0801A586080 2C58E080341BFF07FE00000000000AD7C1E4D3C4C1E5C50000 000000000000
Good luck.
P.S. EBCDIC not ASCII
[ 29 January 2002: Message edited by: bugfix ]
Strike
01-29-2002, 02:08 PM
First of all ... you have to tell us what language this is in (assuming it's a hex dump of assembly, but what arch?).
afaik only the s/390 still uses ebcdic.
YaRness
01-29-2002, 04:22 PM
the answer, of course, is 1s and 0s.
sans-hubris
01-29-2002, 08:16 PM
Is this a game of twenty questions here?
I have to agree with Strike, we need to know the platform. I'd assume it's a big-endian system since the system that uses EBCDIC would be big-endian.
BTW, if anyone's curious, EBCDIC (http://www.everything2.com/index.pl?node=ebcdic).
PS Using EBCDIC makes it that much harder.
[ 29 January 2002: Message edited by: sans-hubris ]
bugfix
01-30-2002, 05:06 AM
Originally posted by jkm:
<STRONG>afaik only the s/390 still uses ebcdic.</STRONG>
Correct, an IBM s/390. I gave EBCDIC as a clue otherwise it would have been near impossible.
And yep, its Assembler.
[ 30 January 2002: Message edited by: bugfix ]
Gnu/Vince
01-30-2002, 08:34 AM
Originally posted by jkm:
<STRONG>afaik only the s/390 still uses ebcdic.</STRONG>
AS/400 too. I would guess then that most IBM mainframes/mini-computers are on EBCDIC