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dhruv · Beginner Joined: 01 Jun 2004 Posts: 10 Topics: 1

I am a little confused with the following scenario

VAR1 : Defined as pic 9(3) Contains low-values (x'000000')
VAR2 : Defined as pic 9(3) contains 0 (x'F0F0F0).

IF VAR1 = ZEROES
( This statement is evaluated as false)

If Var1 = Var2
(This satement is evaluated as true)

Can someone tell me what I am missing here ?

Thanks

kolusu · Posted: Thu Jul 22, 2004 4:30 am Post subject:

dhruv,

how did you populate Low-values to var-1? I don' t think you can have low-values for a numeric integer.

Kolusu
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dhruv · Beginner Joined: 01 Jun 2004 Posts: 10 Topics: 1

Assigned low-values through group level move

Thanks

kolusu · Posted: Thu Jul 22, 2004 4:39 am Post subject:

dhruv,

Please show me your defnition and move statement. I expect a S0C7 abend.

Kolusu
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dhruv · Beginner Joined: 01 Jun 2004 Posts: 10 Topics: 1

There is no SOC7 abend. I am a little surprised that you expect one.

Anyway, what about the following scenario

VAR1 PIC 9(3) .
VAR1_RDF REDEFINES VAR1 PIC X(3).

MOVE LOW-VALUES TO VAR1_RDF

You will not receive a SOC7

In my particular case, its more like this

01 GROUP.
03 FILLER PIC X(1)
...
...
03 VAR1 PIC 9(3).

MOVE LOW_VALUES TO GROUP

kolusu · Posted: Thu Jul 22, 2004 5:05 am Post subject:

dhruv,

I agree that with redefines you will not encounter S0C7 abend. But with a group you will definitely encounter a S0C7 unless there is a redefine.

Kolusu
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Mike Chantrey · Posted: Thu Jul 22, 2004 5:23 am Post subject:

Just going back to the orignal point:
low-values is not regarded as zero in a 9(3) display field (or a packed or floating point field). It is an invalid value. low-values is only regarded as zero in binary field.

dhruv · Beginner Joined: 01 Jun 2004 Posts: 10 Topics: 1

I agree that low-values are binary zeroes.

So If VAR1 = ZEROES (VAR1 = x'000000')
being evluated as false is correct.

But then, why is the statement

IF VAR1 = VAR2 (VAR1 = x'000000' , VAR2 = x'F0F0F0')
being evaluated as true ?

Thanks

slade · Posted: Thu Jul 22, 2004 9:22 am Post subject:

Hi dhruv,

If you change your compiler opts to ...,LIST,NOOFF,... and look at the generated Assembler code for your various instructions, you'll have your answers.

If your REAL ques is "Why did IBM decide to make these instructions behave this way?", only they can answer that. They may even be persuaded to change it, but don't count on it. Mr. Green

Regards, Jack.

dhruv · Beginner Joined: 01 Jun 2004 Posts: 10 Topics: 1

Thanks for all your replies.

The reason i had posted this question was I was under the impression that I was missing something basic.

But if thats the way the cobol compiler is implemented, then ofcourse, i am leaving it at this.

Thanks