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In COBOL CALL-CALLING,if a program A is calling 3 sub-
programs, dynamically, then it is said sub-programs will
always will always in Initial Mode. My question is : Do we
need to code CANCEL or (IS INITIAL) for dynamically called
sub-programs or it is the property of Dynamically called
pgms so every time sub-pgms are called they will be in initial
mode. ***This question is only Dynamic call****,
Please reply. Thank you in advance.

Answers were Sorted based on User's Feedback



In COBOL CALL-CALLING,if a program A is calling 3 sub- programs, dynamically, then it is said sub-p..

Answer / adarsha

The DYNAMIC call has to have 2 load modules separately.
Hence the called module is always in the INITIAL state, you
need not to INITIALIZE or CANCEL.

For STATIC CALL above said things are other way around !

Hope I answered ... lem me know if you have any qustns

Is This Answer Correct ?    5 Yes 0 No

In COBOL CALL-CALLING,if a program A is calling 3 sub- programs, dynamically, then it is said sub-p..

Answer / garry

Thanks Adarsha,

I agree to that. But just wondering why it is said that CANCEL will work only with DYNAMIC Call. If dynamically called PGMs are always in initial mode/state then CAncel has no meaning for dynamic call. I hope u get my confusion regarding this.

Pls reply

Is This Answer Correct ?    1 Yes 0 No

In COBOL CALL-CALLING,if a program A is calling 3 sub- programs, dynamically, then it is said sub-p..

Answer / vinay sonar

NO NEED TO USE CANCLE EVERY TIE.WHEN SUBPROGRAM IS CALLED IT IS ALWAYS IN ITS INITIAL STATE.THIS IS COBOL 85 FEATURE.

Is This Answer Correct ?    1 Yes 1 No

In COBOL CALL-CALLING,if a program A is calling 3 sub- programs, dynamically, then it is said sub-p..

Answer / kingmanish

The dynamically called program is loaded once a call is
made to it dynamically by a calling program.
The program is initialised only when it is called for the
first time and remains in memory till the end of calling
program.

After the first call from second call onwards the called
program should be initialised.

Is This Answer Correct ?    0 Yes 0 No

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