Gods Of COBOL

[Jeremy Falcon]

Might be worthwhile running it through a COBOL to C translator. Then maybe you can make sense of the C code.

Jeremy Falcon

[Douglas Troy]

This is nothing more than a string (alphanumeric) 2816 characters

Chris Losinger wrote:

01 MEMORY-AREA-ONE PIC X(2816).

that second section

Chris Losinger wrote:

01 REDEFINES MEMORY-AREA-ONE. 05 K01-1 OCCURS 37 TIMES INDEXED BY K01-1-X PIC 9(04) COMP-5. 05 K02-1 OCCURS 1369 TIMES INDEXED BY K02-1-X PIC 9(04) COMP-5.

Pic 9(04) COMP-5 is a 4 byte numeric stored in machine independent format Essentially, those two lines redefine the space originally defined by MEMORY-AREA-ONE into two blocks of unsigned 16-bit integers. I cannot recall what the INDEXED allows you to do anymore ... Not sure any of that helps you ... [edit] You might also considering hitting up the COBOL User Groups[^] for help with this, and/or other COBOL related issues you might have. Just a thought.

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:..::. Douglas H. Troy ::..
Bad Astronomy |VCF|wxWidgets|WTL

modified on Friday, April 9, 2010 5:51 PM

[Luc Pattyn]

Hi Chris, you are aware there are a couple of .NET COBOL compilers around? you might be lucky and have a look at the IL your source generates; and reflector might be willing to show you equivalent C# or VB.NET code. And maybe you don't need to translate it after all, just run it... :)

Luc Pattyn [Forum Guidelines] [Why QA sucks] [My Articles]

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Getting an article published on CodeProject should be easier and faster for Bronze and Silver authors.

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[CPallini]

No cookie for you: COBOL is the only computer language invented by the Etrurians. :)

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
[My articles]

[CPallini]

Luc Pattyn wrote:

you are aware there are a couple of .NET COBOL compilers around?

Their authors should be flamed, immediately. :rolleyes:

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
[My articles]

[Steve Mayfield]

Isn't that part of the start up code for Cylons...oh wait, that's "Lords of Kobol" :laugh:

Steve _________________ I C(++) therefore I am

modified on Friday, April 9, 2010 5:55 PM

[martin_hughes]

 01 SORRY-CHRIS                                          PIC YOUR-NAME(LANCELOT).
 
 01 I-DON'T-KNOW.
   05 IDK OCCURS 100 TIMES
          INDEXED BY I-REALLY-DON'T-KNOW                 PIC A-COLOUR(BLUE).
   05 IDK-2 OCCURS 1520 TIMES
          INDEXED BY WHAT-IS-THE-CAPITAL-OF-ASSYRIA      I-DON'T-KNOW-THAT(NONE).  

Output is apparently "Arrrrrrrrrrrrrggggggggggghhhhhhhhhhhhhhhhhh!"

Books written by CP members

[Steve Mayfield]

4 bytes left over Maybe the 4 extra bytes are for the array index variables (K01-1-X & K02-1-X)

Steve _________________ I C(++) therefore I am

[Steve Mayfield]

01 MEMORY-AREA-ONE PIC X(2816).
01 REDEFINES MEMORY-AREA-ONE.
05 K01-1 OCCURS 37 TIMES
INDEXED BY K01-1-X PIC 9(04) COMP-5.
05 K02-1 OCCURS 1369 TIMES
INDEXED BY K02-1-X PIC 9(04) COMP-5.

maybe becomes :confused:

union {
{
PIC MEMORY-AREA-ONE X[2816];
struct {
PIC K01-1-X;
PIC K01-1[37];
PIC K02-1-X;
PIC K02-1[1369];
} COMP-5
};

Steve _________________ I C(++) therefore I am

[JimmyRopes]

Chris Losinger wrote:

'leftover bytes' feels stupid

COBOL programmers were never known to be sticklers for programming best practices. COBOL, like VB, allows non-programmers to think they can program. I have many years of experience in technical support (we programmed in assembler on IBM mainframes) supporting COBOL shops, and I have the scars to prove it.

Simply Elegant Designs JimmyRopes Designs
Think inside the box! ProActive Secure Systems
I'm on-line therefore I am. JimmyRopes

[Chris Losinger]

Douglas Troy wrote:

I cannot recall what the INDEXED allows you to do anymore ..

i think that defines an iterator/pointer that's initialized to the start of the subgroup.

image processing toolkits | batch image processing

[Sean Cundiff]

Chris Losinger wrote:

01 MEMORY-AREA-ONE PIC X(2816).

Allocate a block of memory 2816 bytes in size and fill with 'X'. Name the block of memory 'MEMORY-AREA-ONE'.

Chris Losinger wrote:

01 REDEFINES MEMORY-AREA-ONE.
05 K01-1 OCCURS 37 TIMES
INDEXED BY K01-1-X PIC 9(04) COMP-5.
05 K02-1 OCCURS 1369 TIMES
INDEXED BY K02-1-X PIC 9(04) COMP-5.

Reallocate the array for a different purpose. K01-1 is an array that contains 37 elements. The index to the array is K01-1-X. Each element is a 4-byte numeric. 'PIC 9(04)' Each element is stored as a 2's complement big-endian number in machine-dependant format. 'COMP-5' When I say big-endian I mean that when the number is displayed it will be displayed as 4 digits from left to right with most significant to least significant order. Whether it's big-endian in memory depends on the implementation. K02-1 is a 1369 element array with similar properties. K01-1-X and K02-1-X are the variables used to index the arrays. Each one belongs to it's respective array and can't be used for the other one. p-code example: for K01-1-X in {1 .. 37} do something with K01-1[K01-1-X]

-Sean ---- Fire Nuts

modified on Friday, April 9, 2010 6:58 PM

[Sean Cundiff]

Douglas Troy wrote:

Pic 9(04) COMP-5 is a 4 byte numeric stored in machine independent format

Machine-dependent format if I'm not mistaken. COMP-4 is machine independent.

-Sean ---- Fire Nuts

[Chris Losinger]

how can they be 4 byte integers ? that's twice the size of the MEMORY-AREA-ONE's 2816 bytes. (37 + 1369) * 4 = 5624 i'm thinking the 4 is the number of places allowed for the value. 9(4) is a 4 digit integer (0-9999) ? you can fit that in a two byte integer, which would fit in the 2816 bytes. make sense ?

image processing toolkits | batch image processing

[Sean Cundiff]

yeah I updated my post after I realized I put the wrong info in it. 4 byte 'numeric'. the numeric is stored in memory as a 2's complement number, big-endian and machine-dependent format. So 37 * 2 + 2 for the first array, and 1369 * 2 + 2 for the second array. ==> 76 + 2740 = 2816 bytes.

-Sean ---- Fire Nuts

[Dan Mos]

It throws an exception!! Either a System_TerrorException or a System_LiquidNitro one.

[AspDotNetDev]

Or a Dumber_Than_Baboon_Exception.

[Forum Guidelines]

[Dan Mos]

My bad. I totally forgat about it. Forgime me pls!!! URGENT!!!

[BonshatS]

I knew cobol in another long ago and far away life.

01 MEMORY-AREA-ONE PIC X(2816).

01 REDEFINES MEMORY-AREA-ONE.
05 K01-1 OCCURS 37 TIMES
INDEXED BY K01-1-X PIC 9(04) COMP-5.
05 K02-1 OCCURS 1369 TIMES
INDEXED BY K02-1-X PIC 9(04) COMP-5.

If I remember right, that corresponds to :

struct numeric_arrays
{
short k01_1_x;
short k01_1[37];
short k02_1_x;
short k02_1[1369];
};

union mem_area
{
char memory_area_one[2816];
numeric_arrays k01_and_k02;
};

As others have said, the PIC 9(04) is 4 numeric digits and the comp-5 packs it to 2 bytes. I believe the indices map to mem preceding the array parts in mem, but I'm not 100% sure.

[chrissb]

Hahaha. Excellent dude. Best Holy Grail reference I've seen yet.