Zero space allocation is perfectly valid. As is SPACE (0,1) also. The
result is just as requested. In either case, the data set exists in the
VTOC but takes up no space on disk. The data set is treated as 'real',
including GRS enqueue. Hence it can be used like any other exclusively
held data set to serialize execution.
.
.
JO.Skip Robinson
SCE Infrastructure Technology Services
Electric Dragon Team Paddler
SHARE MVS Program Co-Manager
626-302-7535 Office
323-715-0595 Mobile
jo.skip.robin...@sce.com
From: Paul Gilmartin <paulgboul...@aim.com>
To: IBM-MAIN@LISTSERV.UA.EDU
Date: 08/25/2012 08:27 AM
Subject: Re: Space Allocation In Bytes
Sent by: IBM Mainframe Discussion List <IBM-MAIN@LISTSERV.UA.EDU>
On Wed, 8 Aug 2012 14:06:04 +0200, R.S. wrote:
>
> .... BTW: your allocation request was
> illogical - you wanted to have 80-byte records and requested 1 byte.
> Such request has to be re-interpreted or canceled. ;-)
>
As in:
DD LRECL=80,SPACE=(1,...),...
Seveal contributors argued that there was no way a 1-byte block
could be written if LRECL=80, and the construct should result in
an error.
The JCL RM says:
blklgth -- (only if AVGREC is not coded)
Specifies the average block length of the data, in bytes.
The blklgth is a decimal number from 0 through 65535.
Really!? In fact by experiment, in JCL:
DD SPACE=(0,1),...
and in TSO
ALLOCATE AVBLOCK(0) ...
are both accepted without complaint. I suppose allocation adds a
count and an IBG; divides track size by that; takes the ceiling and
requests the resulting number of tracks (almost certainly 1). I have
little problem with that. I haven't investigated whether SPACE=(0,99999)
allocates fewer tracks than SPACE=(1,99999). It's possible that integer
arithmetic or 32-byte chunking gives the same result for both.
Perhaps I'll start coding SPACE=(0,1) in JCL to allocate minimal
data sets, just to startle readers.
-- gil
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