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How to caliculate Working storage length


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lokanathareddy

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Joined: 05 Mar 2005
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PostPosted: Mon Oct 20, 2008 5:27 pm
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Hi,

I have a requirement with Working storage length. I want to caliculate whole working storage length (All variables). I have heard that we can caliculate thru BLW address

Please can some one eloborate to understand in easy way.
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Sinkaravelan S

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Joined: 21 Apr 2008
Posts: 27
Location: coimbatore-TN india

PostPosted: Mon Oct 20, 2008 6:14 pm
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lokan,

one of the easiest way is copy all the working storage variable into one ps...

afterthat,open file aid an select view option( option 8 in file-aid)..

give your ps name, it will display all the variable storage length andalso we can know the total length of all the variables.


Thanks,
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Robert Sample

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PostPosted: Mon Oct 20, 2008 11:55 pm
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Check the Application Programming Guide for the MAP option -- which produces a Data Division map including the BLW cell number and offset. Looking at the last item in the map gives you the total BLW cells used and the offset in the last one. Since each BLW cell is 4K, it is pretty simple to calculate total length of working storage.
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Anuj Dhawan

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Joined: 22 Apr 2006
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PostPosted: Tue Oct 21, 2008 3:19 am
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Hi,

This link might also provide you a start..

www.ibmmainframes.com/viewtopic.php?t=35113&highlight=fileaid
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Escapa

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PostPosted: Tue Oct 21, 2008 11:03 am
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In compiler listing i found below
When i defined on only one variable of X(20) i.e. HEX 14..

Can anybody flash more light on this?
Is the bolded line showing length of working storage section.

TGT WILL BE ALLOCATED FOR 00000148 BYTES
SPEC-REG WILL BE ALLOCATED FOR 0000007E BYTES
WRK-STOR WILL BE ALLOCATED FOR 00000014 BYTES
DSA WILL BE ALLOCATED FOR 000000F8 BYTES
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Robert Sample

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PostPosted: Tue Oct 21, 2008 4:38 pm
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Hex 0014 = 1 x 16 + 4 = 20 bytes
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Anuj Dhawan

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PostPosted: Wed Oct 22, 2008 1:44 am
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That, shown in bold, is HEX equivalemt of 20, Compiler is working well at your end.. icon_wink.gif

You should not get confused, if there is written "00000014", if in
Quote:
SPEC-REG WILL BE ALLOCATED FOR 0000007E BYTES
"0000007E " is HEX for sure, then, in general, other similar description should also be in HEX...umm..just trying to be rather simple..
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Escapa

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PostPosted: Wed Oct 22, 2008 11:03 am
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Hi Anuj, It is surely in HEX..
Because if i use one variable with x(40)
it gave following..Where 28 hex is 40 dec.
Quote:

TGT WILL BE ALLOCATED FOR 00000148 BYTES
SPEC-REG WILL BE ALLOCATED FOR 0000007E BYTES
WRK-STOR WILL BE ALLOCATED FOR 00000028 BYTES
DSA WILL BE ALLOCATED FOR 000000F8 BYTES


But one doubt i have is if i use two variables of x(20) each then it shows 2C which is 44
Quote:

TGT WILL BE ALLOCATED FOR 00000148 BYTES
SPEC-REG WILL BE ALLOCATED FOR 0000007E BYTES
WRK-STOR WILL BE ALLOCATED FOR 0000002C BYTES
DSA WILL BE ALLOCATED FOR 000000F8 BYTES


But one doubt i have is if i use three variables of x(20) each then it shows 44 which is 68
Quote:

TGT WILL BE ALLOCATED FOR 00000148 BYTES
SPEC-REG WILL BE ALLOCATED FOR 0000007E BYTES
WRK-STOR WILL BE ALLOCATED FOR 00000044 BYTES
DSA WILL BE ALLOCATED FOR 000000F8 BYTES


Can anybody clarify on this?
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dbzTHEdinosauer

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PostPosted: Wed Oct 22, 2008 11:20 am
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are your variables defined as level > 01 or as level 01?
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Escapa

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PostPosted: Wed Oct 22, 2008 11:23 am
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Quote:
are your variables defined as level > 01 or as level 01?

All are at 01 level
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Robert Sample

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PostPosted: Wed Oct 22, 2008 4:43 pm
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From the Language Reference manual section 5.3.15:
Quote:
In the working-storage section, the compiler aligns all level-01 entries on a doubleword boundary.
So your first variable is followed by a 4-byte filler to force alignment to a doubleword; the second variable is followed by a 4-byte filler to force alignment to a doubleword.

20 + 4 + 20 + 4 + 20 = 68
The computer doesn't lie -- you just have to understand what it tells you.
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Escapa

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PostPosted: Wed Oct 22, 2008 5:42 pm
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Thanks Robert.
When i defined three variables of x(24) it correctly shows hex 48 i.e. 24*3=72 in dec as there is no need to put filler to force alignment its already icon_smile.gif icon_smile.gif

Quote:

TGT WILL BE ALLOCATED FOR 00000148 BYTES
SPEC-REG WILL BE ALLOCATED FOR 0000007E BYTES
WRK-STOR WILL BE ALLOCATED FOR 00000048 BYTES
DSA WILL BE ALLOCATED FOR 000000F8 BYTES

I read the details at
publib.boulder.ibm.com/infocenter/ratdevz/v7r1m1/topic/com.ibm.ent.cbl.zos.doc/topics/igy3lr31.pdf

Thanks a lot again.
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pkmurali
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Joined: 15 Dec 2005
Posts: 271

PostPosted: Thu Sep 02, 2010 12:43 pm
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Hi all,

I have the same scenario what lokanathareddy has faced. my module contains only one 01 level
as shown below,


Code:

           01 ws-stor-var          pic s9(04) comp.



The size shown in listing is


Code:

TGT         WILL BE ALLOCATED FOR 00000150 BYTES
SPEC-REG WILL BE ALLOCATED FOR 0000007E BYTES
WRK-STOR WILL BE ALLOCATED FOR 00000654 BYTES
DSA      WILL BE ALLOCATED FOR 00000110 BYTES


Please let me know how the size of s9(04) comp is related as 654 bytes?
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pkmurali
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Posts: 271

PostPosted: Thu Sep 02, 2010 3:11 pm
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I am sorry, my compiler has some more options which leads to extra 652 bytes. It is due to s9(04) comp (2 bytes + 4 slack bytes due to double word+294 bytes due to sql-vers +354 bytes due defheil ...)

sorry for the confusion.

thanks,
Murali.
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