Re: RCVF in CL

On 10-Feb-2015 09:20 -0600, rob@xxxxxxxxx wrote:

Normally in the second program concept you do not have two programs
reading the file. It's often done like this:

myloop:
call ReaderProgram
dlyjob 300
goto myloop
endpgm

<<SNIP>>

Not in my experience. The implication of someone suggesting using two programs had been typically, to suggest the use of two distinct programs whereby each had coded both the identical DCLF and identical looping through the data with RCVF [terminating upon CPF0864], but each program performs different work against the data than the other.

The above-quoted CL example reflects what I expect would be described as having coded just _one program_ that accepts a parameter to describe what that one program does; the program would be called twice, and the parameter tells the program which _pass_ of the data, such that even if the caller were understood to be a program [rather than scripted calls], then that program would not be /counted/.

And what is shown in that quoted example, fails to explain how the ReaderProgram knows whether each particular invocation should perform the pass-one or pass-two.? I would expect that if someone were to respond with a two-program suggestion whereby only one does the DCLF, then they would be suggesting instead, something like the following?:

mainline: /* program: InvokeReaderProgram */
call ReaderProgram parm('*PASS1')
call ReaderProgram parm('*PASS2')
endpgm