I am trying to convert a file that has packed binary fields. The files were created on a S36, and now reside on an AS400. I am trying to write a program that takes those fields and writes to another file that those fields are uncomped. When I run my program I keep getting file accessed with wrong record length. Is there a special way in cobol to handle those kinds of fields. Thanks in advance for any help anyone can provide me.
Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations SkipVought on being selected by the Tek-Tips community for having the most helpful posts in the forums last week. Way to Go!
Binary Fields
- Thread starter GaryS2002
- Start date
- Thread starter
- #3
Crox,
Thanks for replying. Here is what I have. These files were created using RPG. With RPG there is a code you can use that declares a field or array as binary packed. This file I am working with declares an array as 20 entries at size 7, it is however binary packed. Now I know that a straight binary is half of the actual size, and you declare it as a comp field in coobol. But when it is packed do I declare it as comp-3 and take half + 1? If you need more info let me know nad I will get the file specs. Thanks alot for all your help.
Thanks for replying. Here is what I have. These files were created using RPG. With RPG there is a code you can use that declares a field or array as binary packed. This file I am working with declares an array as 20 entries at size 7, it is however binary packed. Now I know that a straight binary is half of the actual size, and you declare it as a comp field in coobol. But when it is packed do I declare it as comp-3 and take half + 1? If you need more info let me know nad I will get the file specs. Thanks alot for all your help.
- Thread starter
- #5
On my RPG program print out. I have an extension that declares a field name, occurs clause, size of entry, and decimal position, looking like this
E 20 7 0 VL
Then when I get to the I specification
I P 215 294 VL
Now when you take and figure 215-294+1 you get 80, but when you look at the extension specification 20*7 you get 140. My problem is how do I declare the elements from the I specification in cobol? I did this I used comp-3 taking 12/2+1 which would get 7 then took that 20 times. When I compiled my program with xref it reported my file length as 500 somthing. It needs to be 384.
E 20 7 0 VL
Then when I get to the I specification
I P 215 294 VL
Now when you take and figure 215-294+1 you get 80, but when you look at the extension specification 20*7 you get 140. My problem is how do I declare the elements from the I specification in cobol? I did this I used comp-3 taking 12/2+1 which would get 7 then took that 20 times. When I compiled my program with xref it reported my file length as 500 somthing. It needs to be 384.
I am no great RPG expert -- fair warning. ![[glasses]](/data/assets/smilies/glasses.gif "[glasses] [glasses]")
80 / 20 = 4. So there seem to be 4 character positions used to represent 7 digits. But this is exactly the number of character positions that a packed decimal (COMP-3) item would take. So I would think that you should be using a COBOL description of (for example):
I am not sure where the 12 came from in your "12/2+1" expression. Tom Morrison
80 / 20 = 4. So there seem to be 4 character positions used to represent 7 digits. But this is exactly the number of character positions that a packed decimal (COMP-3) item would take. So I would think that you should be using a COBOL description of (for example):
Code:
03 my-number PIC S9(7) COMP-3 OCCURS 20.- Thread starter
- #7
- Status
Similar threads
