Re: Replication factor of a group item

2014-11-26 Thread Micheal Butz
Wow I get a reply from the author himself I'll try it thanks Sent from my iPhone > On Nov 26, 2014, at 8:16 PM, John Ehrman wrote: > > Micheal, Could you write DS 0CL(11*133) ? > > Regards... John Ehrman

Re: Replication factor of a group item

2014-11-26 Thread John Ehrman
Micheal, Could you write DS 0CL(11*133) ? Regards... John Ehrman

Re: Replication factor of a group item

2014-11-26 Thread Micheal Butz
The problem is that file manager see 2 separate dsects as as 2 records Sent from my iPhone > On Nov 26, 2014, at 6:10 PM, John Gilmore wrote: > > The '0' in > > 0CL133 > > is called a duplication factor. It is not called a replication factor. > > Writing > > 11CL133 > > thus specifies 11

Re: Replication factor of a group item

2014-11-26 Thread John Gilmore
The '0' in 0CL133 is called a duplication factor. It is not called a replication factor. Writing 11CL133 thus specifies 11 instances of 133 characters, and any duplication factor of 0 specifies no instances, only alignment, so that 0H specifies halfword alignment, 0F specifies fullword alignm

Re: Replication factor of a group item

2014-11-26 Thread Micheal Butz
I guess for lack of a better forum (as this dsect is being parsed by file manager) it sees 0CL133 as a group item I am trying to create any array Of items which equal 133 Sent from my iPhone > On Nov 26, 2014, at 5:22 PM, retired mainframer > wrote: > > What is the desired difference to eith

Re: Replication factor of a group item

2014-11-26 Thread retired mainframer
What is the desired difference to either the assembler or your object code between DS 0CL133 And DS 11 * 0CL133 if that were allowed? In what way does 11 instances of nothing differ from 1instance or no instance? > -Original Message- > From: IBM Mainframe Assembler List [mailto:

Re: Replication factor of a group item

2014-11-26 Thread Micheal Butz
You are correct however file manager in addition to alignment Takes 0cl As a group item Was wondering if there is anyway to code an array of a group item Sent from my iPhone > On Nov 26, 2014, at 4:30 PM, Rob van der Heij wrote: > > What is 11 times "no string of 33 characters" ? Are you

Re: Replication factor of a group item

2014-11-26 Thread Rob van der Heij
What is 11 times "no string of 33 characters" ? Are you sure you need 11 empty strings rather than 10 ? nnCL33 is "nn" times a field of type C and length 33. When "nn" is 0 you don't actually allocate but only align. Which is why you see 0F or 0D. But for type "C" there is no alignment... On 26

Replication factor of a group item

2014-11-26 Thread Micheal Butz
Hi This is really a follow up to my file manager question I would like to define a array of a group item So I coded for example 110cl33 Where I meant 11 x 0cl33 However for the assembler it read it as 110 x 33 Is there a way to code 11 x 0cl33 Thanks Sent from my iPhone

Re: ASSEMBLER-LIST Digest - 23 Nov 2014 to 25 Nov 2014 (#2014-62)

2014-11-26 Thread John Walker
You would just allocate some memory and then save the pointer to it. The maximum size of that amount would be the upper limit. You would keep a pointer to the last known 'end' for the rest of it. Pretty basic. On Tue, 11/25/14, ASSEMBLER-LIST automa

Re: File manger arrays in assembler

2014-11-26 Thread Micheal Butz
It's not cobol the copybook/dsect is assembler I am hoping that when I hit the array section of the record I'm able to pf8/ page forward Sent from my iPhone > On Nov 26, 2014, at 2:12 AM, Dougie Lawson wrote: > > Hi Michael, > > What does your COBOL copybook look like? > > The usual way is