CODEWITHSUNDEEP
cobolgnucobolcobol85
user99999991
user99999991

Reputation: 1431

Call function in COBOL more than once

I have my main calling two functions. The second function called (Decrypt) calls the first function (Encrypt) inside of it. So here Encrypt is being called twice. Once in the main, and then once inside of Decrypt.

The issue is that it refuses to work this way. Once Encrypt gets used in the main, I can't use Encrypt again anywhere~ in the program. Its like the variables are still in use and I can't pass it new ones.

For example, if I remove Encrypt from the main function and ONLY call Decrypt - it works fine. I can't figure out why.

IDENTIFICATION DIVISION.
PROGRAM-ID. CAESER-1-CIPHER.
DATA DIVISION.
PROCEDURE DIVISION
CALL 'ENCRYPT' USING BY CONTENT INPUTE CIPHERE.
CALL 'DECRYPT' USING BY CONTENT INPUTD CIPHERD.
STOP RUN.

IDENTIFICATION DIVISION.
PROGRAM-ID. ENCRYPT.
DATA DIVISION.
PROCEDURE DIVISION BLAH BLAH
BLAH BLAH COMPUTE
END PROGRAM ENCRYPT.

IDENTIFICATION DIVISION.
PROGRAM-ID. DECRYPT.
DATA DIVISION.
PROCEDURE DIVISION BLAH BLAH
CALL 'ENCRYPT' USING BY CONTENT BLAH BLAH
EXIT PROGRAM.
END PROGRAM DECRYPT.

Upvotes: 1

Views: 2149

Answers (3)

Bill Woodger
Bill Woodger

Reputation: 13076

If your compiler supports it, look at the LOCAL-STORAGE SECTION.

This will "automtically" set all values to initial state each time the sub-program is called.

They are sub-programs, not functions.

NealB's suggestion of IS INITIAL will work. The advantage of the LOCAL-STORAGE is when you have a "mixture" of things: some whose value you want retained between CALLs, for which you use WORKING-STORAGE, and others which you specifically want to have reset to an initial state, which you define in the LOCAL-STORAGE, with VALUE clauses.

The processing that IS INITIAL does is equivalent to the processing for the LOCAL-STORAGE. Everything in the WORKING-STORAGE which has a VALUE clause is set to that value each time the program-name is CALLed. Fields without VALUE clauses have an "undefined" content.

Upvotes: 2

NealB
NealB

Reputation: 16928

I'm not sure I follow your question completely, I would have expeced more problems with DECRYPT. This is why...

Program SER-1-CIPHER contains two nested programs: ENCRYPT and DECRYPT. Until you declare ENCRYPT and DECRYPT as COMMON programs they cannot "see" each other because only programs at higher levels of nesting (eg. SER-1-CIPHER) can "see" programs that are nested within them. This is explained in the Open Cobol Programmers Guide on nested programs.

Try declaring nested programs as:

 PROGRAM-ID. ENCRYPT IS COMMON.
 PROGRAM-ID. DECRYPT IS COMMON.

This way program DECRYPT will be able to CALL ENCRYPT.

Next, I would encourage you to use GOBACK in place of STOP RUN and EXIT PROGRAM when returning control to the operating system or calling programs. The OpenCobol Programmers Guide also makes this recommendation.

Finally, each subprogram should contain as its last statement a GOBACK. I am not sure what the behaviour of ENCRYPT is without having an explicit return statement of some kind. Your actual program may have this, but the code example in your question doesn't.

Open Cobol does not seem to have any restrictions on having recursion among nested programs, but some versions of COBOL do not allow this (eg. IBM Enterprise COBOL).

Upvotes: 1

user99999991
user99999991

Reputation: 1431

The data instantiated in the encrypt function is still left over. They need to be reset. So any number PICs in Encrypt needed to be set to 0 at the beginning of the function otherwise they would still have data stored in them.

Example:

COMPUTE MYNUM1 = 0
COMPUTE MYNUM2 = 0
COMPUTE MYNUM3 = 0
MOVE '' TO MYVARSTRING

Upvotes: 0

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