OW0007OB is only 2 bytes followed by 2 unused bytes.

The reason is that OQ0007AC is defines as a full word and thus has to be
full word aligned at address 000804.  If you had defined OW0007AC as just
XL4 it would've started at 000802.  Similarly, if you had defined it as D,
then it would have to be double word aligned and been located at 000808.




Have a nice day,
Dave Betten
z/OS Performance Specialist
Cloud and Systems Performance
IBM Corporation
email:  [email protected]


IBM Mainframe Discussion List <[email protected]> wrote on
07/04/2020 03:48:03 PM:

> From: Nguyen Dt <[email protected]>
> To: [email protected]
> Date: 07/04/2020 03:48 PM
> Subject: [EXTERNAL] Assembler question
> Sent by: IBM Mainframe Discussion List <[email protected]>
>
> Dear lister,
> I am learning assembler on my own, i have something strange that i
> can't explain , please help me to understand
>
> Here is a section of my code :
>
> 608 DBLWORD  DS  D                DBLE WORD
>
> 609 PATTERN6 DC  X'402020202120'
> 610 *
>
> 611 REPORTO7 DS  0CL274
>
> 612 OW0007DB DS  XL2
>
> 613 OW0007OB DS  XL2
>
> 614 OW0007AC DS  F
>
> 615 OW0007NP DS  H
>
> 616 OW0007PT DS  F
>
> 617 OW0007PF DS  CL256
>
> 618 *
>
>
>
> And here is the listing at compilation
>
> 0007FE                              611 REPORTO7 DS  0CL274
>
> 0007FE                              612 OW0007DB DS  XL2
>
> 000800                              613 OW0007OB DS  XL2
>
> 000804                              614 OW0007AC DS  F
>
> 000808                              615 OW0007NP DS  H
>
> 00080C                              616 OW0007PT DS  F
>
>
>
> I don't undestand why OW0007OB is 2 bytes  , but on the left it is
> considered as 4 bytes long , the same for QW0007NP
> I read something with alignment obut it seems to concern constants ,
> and this is not constant ?
>
>
> Thank you for clarification
>
>
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