Just for fun, I tried another quick 6.3 test using the Repository setup.
Here are the nested function results:
Function Length(Function Trim(indexed field)) - Worked fine
Function Length( Trim(indexed field)) - Worked fine
Length(Function Trim(indexed field)) - Failed - only used
indexedfield(1)
Length( Trim(indexed field)) - Failed - only used
indexedfield(1)
So it seems the problem seems to be with the Length function, when
called using the Repository Intrinsic functions description.
Got to get this programmer back in the saddle, so that's all I have time
for here. Thanks again!
Billy
------ Original Message ------
From "Billy Ashton" <[email protected]>
To "IBM Mainframe Discussion List" <[email protected]>
Date 5/24/2023 1:24:05 PM
Subject Re[2]: COBOL Field Length problem
Michael, you gave me something interesting to try. It seems the Configuration
Section use of the Repository for Intrinsic functions is new with 6.3, as it
did not compile under 6.2.
So...I removed that paragraph and added the explicit references to the FUNCTION
keyword and got the same results that you did - everything worked fine. I then
moved up to 6.3 with that working code, and guess what? It worked there, too.
So it seems that there is some problem with how that Repository Intrinsic
function thing is not working, and I haven't the time to chase it down right
now. I will just use explicit FUNCTION calls to get past the problem.
Thanks for pointing me the right direction through the fog!
Billy Ashton
------ Original Message ------
From "Schmitt, Michael" <[email protected]>
To [email protected]
Date 5/24/2023 1:03:18 PM
Subject Re: COBOL Field Length problem
I don't have 6.3, but I tried your program on 6.2 after adding FUNCTION before
the TRIM and LENGTH functions.
The results seem to be as expected. For example:
00000005 ----+----1----+----2----+----3----+----4----+----5----+----6----+
Original: > COMND VALUE3 21 <
Trim: > COMND VALUE3 21<
Lengths: I1(00000021) I2(00000021) Hold(00000021) By name(00000021)
00000006 ----+----1----+----2----+----3----+----4----+----5----+----6----+
Original: >COMND VALUE4 15 <
Trim: >COMND VALUE4 15<
Lengths: I1(00000015) I2(00000015) Hold(00000015) By name(00000015)
I'd suggest reducing the test case down to the MINIMUM amount of code that
demonstrates the issue.
Then look at the p-map to see what the compiler is doing.
-----Original Message-----
From: IBM Mainframe Discussion List <[email protected]> On Behalf Of
Billy Ashton
Sent: Wednesday, May 24, 2023 11:25 AM
To: [email protected]
Subject: COBOL Field Length problem
Hey everyone - back again with another COBOL problem. Your help with my
COPY REPLACING question was great and the programmer is quite happy with
that solution. Now, he came to me with what looks like a problem, and I
am not sure if we are doing something wrong, or if it is a bug in our
Enterprise COBOL for z/OS 6.3.0 P220314 system.
In a nutshell, when using the LENGTH(TRIM(some-field)) function against
any elementary data item, it works great. However, when using it against
an item within an occurs (think data table), every reference beyond 1
gets handled as item #1. For example, if I have a group with 8 items,
the length of item #1 is right, but the length of item(2) through
item(8) is always the value of item(1). The table index can be display
numeric, packed, or binary, and the results are the same, so I don't
think it is a problem with the index, but somehow the reference is not
resolved correctly within the nested function.
Maybe a short program would be helpful. I hope a 60 line program is ok!
Let me know what you think is happening.
IDENTIFICATION DIVISION.
PROGRAM-ID. TSTPG002.
ENVIRONMENT DIVISION.
Configuration Section.
Repository.
Function All Intrinsic.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 WS.
05 INL-NO PIC S9(08) VALUE ZERO BINARY.
05 INL-I1 PIC S9(08) VALUE ZERO BINARY.
05 INL-I2 PIC S9(08) VALUE ZERO BINARY.
05 INL-H PIC S9(08) VALUE ZERO BINARY.
05 IN-GRP-X.
10 L1 PIC X(65) VALUE '* THIS_IS_A_COMMENT Here .28'.
10 L2 PIC X(65) VALUE SPACE.
10 L3 PIC X(65) VALUE 'COMND VALUE1 17'.
10 L4 PIC X(65) VALUE ' COMND VALUE2 21'.
10 L5 PIC X(65) VALUE ' COMND VALUE3 21'.
10 L6 PIC X(65) VALUE 'COMND VALUE4 15 '.
10 L7 PIC X(65) VALUE ' COMND VAL* 27 '.
10 L8 PIC X(65) VALUE ' * THIS_IS_A_COMMENT... 29'.
05 REDEFINES IN-GRP-X OCCURS 8.
10 IN-LINE PIC X(65).
05 Hold-L PIC X(65).
05 I1 PIC S9(08) VALUE ZERO Binary.
05 I2 PIC S9(08) VALUE ZERO.
PROCEDURE DIVISION.
PERFORM VARYING I1 FROM 1 BY 1 UNTIL I1 GREATER 8
Move I1 to I2
Move In-line(I1) to Hold-L
Display I1 ' '
'----+----1----+----2----+----3----+----4'
'----+----5----+----6----+'
Display ' Original: >' IN-LINE(I1) '<'
Display ' Trim: >' Trim(In-line(I1) Trailing) '<'
Compute INL-I1 = Length(Trim(In-line(I1) Trailing))
Compute INL-I2 = Length(Trim(In-line(I2) Trailing))
Compute INL-H = Length(Trim(Hold-L Trailing))
Evaluate TRUE
When I1 = 1 Compute INL-NO = Length(Trim(L1 Trailing))
When I1 = 2 Compute INL-NO = Length(Trim(L2 Trailing))
When I1 = 3 Compute INL-NO = Length(Trim(L3 Trailing))
When I1 = 4 Compute INL-NO = Length(Trim(L4 Trailing))
When I1 = 5 Compute INL-NO = Length(Trim(L5 Trailing))
When I1 = 6 Compute INL-NO = Length(Trim(L6 Trailing))
When I1 = 7 Compute INL-NO = Length(Trim(L7 Trailing))
When Other Compute INL-NO = Length(Trim(L8 Trailing))
End-Evaluate
Display ' Lengths:'
' I1(' INL-I1 ')'
' I2(' INL-I2 ')'
' Hold(' INL-H ')'
' By name(' INL-NO ')'
Display ' '
END-PERFORM
GOBACK.
Thank you and best regards,
Billy Ashton
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