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ashokkit
New User
Joined: 22 Apr 2005 Posts: 4
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01 rec1 .
02 a pic 9(3) value 222.
02 b pic 9(2) value 11.
01 rec2 redefines rec1 .
02 c pic 9(4).
02 filler pic x.
??
procedure division.
?????display c.
what is the anwer for this?
[/i] |
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mmwife
Super Moderator
Joined: 30 May 2003 Posts: 1592
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Hi,
The ans is 2221 |
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anuradha
Active User
Joined: 06 Jan 2004 Posts: 247 Location: Hyderabad
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Hi Ashok!
A small suggestion from me. Why don't you give a simple try with that code, rather than asking us. Experience teaches us many things and even we get a chance to expermient many things with that piece of code right!
Hope you will take my suggestion in a positive sense. |
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ashokkit
New User
Joined: 22 Apr 2005 Posts: 4
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hai thanks alot . i got a little bit of confusion .i thought that redefines will not occupy values but uses the storage space alone . thats y ? i got clarified now . |
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mmwife
Super Moderator
Joined: 30 May 2003 Posts: 1592
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Hi Ashok,
I'm not sure I get your point, but I'll explain my understanding:
When you define a group field w/ valued elementary sub-fields they define a string of data. In your example that string is 5 bytes long and contains "22211":
a - 3 bytes long and containing "222".
b - 2 bytes long and containing "11".
When you redefined that 5 byte string you divided it into 2 fields:
A 4 byte field (c) that contained the 1st 4 bytes of the string, "2221".
A 1 byte field (filler) that contained the 5th byte of the string, "1".
HTH |
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