>>With this change, when the JCL interpreter sees the PARMDD=, it >>allocates the buffer of 32768 bytes > >Did you see that documented somewhere?
As one should expect, how the system chooses to allocate storage it uses is not documented, as there is no need for the user to care. Informally, for those curious, the processing does what you might expect it to do. It allocates system subpool storage of approximately the maximum size so that the PARMDD data can be read in, up to the maximum size. When the parm data is to be presented to the user, an area of the needed size, in a user subpool, is used, copied from the system storage. What I don't recall anyone inquiring about is -- what columns of the PARMDD data are used? Answer: all, unless for fixed-block the last 8 characters are decimal-numeric in which case those are treated as a sequence number and are stripped out -- what about trailing blanks (including those preceding a stripped out sequence number)? Answer: they are stripped out -- in the result, where does the first column of line N+1 land in the resulting parameter string with respect to the last-used column of line N? Answer: right next to it (no extra blanks inserted in between lines) -- what about leading blanks? Answer: they are kept. So if you need a blank separating what is on line N and line N+1, you would start line N+1 with a blank Peter Relson z/OS Core Technology Design ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to [email protected] with the message: INFO IBM-MAIN
