RE: "If %scan(x:y)" is not valid???

That's why I've seen people asking for built-in functions like:

c     if    acflag = %list( 'a' 'b' 'c' )

or

c     if    date = %range( 030101 030201 )


For reading sake you could always:

d notFound       s         


   C          IF           ACFLAG = 'A'
   C                         AND %scan( W_WIPSTA : 'BHIP') <> notFound
   C                         AND %scan( W_BILTYP : 'CFR' ) > 0

-----Original Message-----
From: Dan [mailto:dbcerpg@yahoo.com]
Sent: Monday, February 10, 2003 3:55 PM
To: rpg400-l@midrange.com
Subject: "If %scan(x:y)" is not valid???


(All variable fields described here are 1-character)

I was somewhat surprised to find that 
   C          IF        %scan( ACFLAG : 'ABC' ) 
does not compile, even at v5r2.  I have to add a "> 0" test to make it work,
i.e., 
   C          IF        %scan( ACFLAG : 'ABC' ) > 0

I am trying to avoid the 
   C  ACFLAG  SCAN   'ABC'
   C          IF     %FOUND

Currently, I am looking at a block of code like this
   C          IF           ACFLAG = 'A'
   C                         AND (   W_WIPSTA = 'B'
   C                              OR W_WIPSTA = 'H'
   C                              OR W_WIPSTA = 'I'
   C                              OR W_WIPSTA = 'P')
   C                         AND (   W_BILTYP = 'C'
   C                              OR W_BILTYP = 'F'
   C                              OR W_BILTYP = 'R')
and would like to simplify it thusly:
   C          IF           ACFLAG = 'A'
   C                         AND %scan( W_WIPSTA : 'BHIP') > 0
   C                         AND %scan( W_BILTYP : 'CFR' ) > 0

(Actually, my example is simplified; I have more tests than that.)
Is this understandable?  At first glance, I found it confusing.  I would
much rather code this
more like:
   C          IF           ACFLAG = 'A'
   C                         AND W_WIPSTA = ('B' 'H' 'I' 'P')
   C                         AND W_BILTYP = ('C' 'F' 'R')
where multiple values in a parentheses set in an "if equal" test logically
implies an OR grouping.

Curious minds need to know...
TIA, Dan

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