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Usage of RANDOM function in COBOL


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muthuvel

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Joined: 29 Nov 2005
Posts: 217
Location: Canada

PostPosted: Sun Nov 05, 2006 7:37 pm
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Hi,
Has anybody used RANDOM function in COBOl?

If so,Can you please explain me the function with a piece of batch code?

I went through the manual but it is easy to understand with a piece of code.

Thanks in Advance,

Thanks,
Muthuvel.
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priyesh.agrawal

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Joined: 28 Mar 2005
Posts: 1448
Location: Chicago, IL

PostPosted: Mon Nov 06, 2006 3:04 am
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The random function takes a non-negative integer argument and returns a decimal number. The integer argument is optional. It is the seed for a pseudo-random mathematical process, that determines the function’s first returned number. In computer languages, all random processes are in fact long, repeating series of numbers that appear to be random. A well-chosen process repeats only after billions of values, so it is a practical source of random numbers, even if it is not a theoretically pure one. If you supply the random function with an argument, it will generate the same sequence of numbers. Without the argument, the random function generates the next number in its sequence.

Quote:
I went through the manual but it is easy to understand with a piece of code.

Could you put the manual link describing RANDOM FUNCTION.
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muthuvel

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Joined: 29 Nov 2005
Posts: 217
Location: Canada

PostPosted: Wed Nov 08, 2006 2:32 pm
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Priyesh,
This is the lengthy stuff :? which i got for RANDOM function in COBOL.

The RANDOM function returns a numeric value that is a pseudorandom number
from a rectangular distribution.

The function type is numeric.

+--- Format -------------------------------------------------------------+
| |
| >>--FUNCTION RANDOM------------------------------------------------->< |
| +-(argument-1)-+ |
| |
+------------------------------------------------------------------------+
argument-1
If argument-1 is specified, it must be zero or a positive integer, up
to and including (10**18)-1 which is the maximum value that can be
specified in a PIC9(18) fixed item; however, only those in the range
from zero up to and including 2,147,483,645 will yield a distinct
sequence of pseudorandom numbers.

If a subsequent reference specifies argument-1, a new sequence of
pseudorandom numbers is started.

If the first reference to this function in the run unit does not specify
argument-1, the seed value used will be zero.

In each case, subsequent references without specifying argument-1 return
the next number in the current sequence.

The returned value is exclusively between zero and one.

For a given seed value, the sequence of pseudorandom numbers will always be the same.



Regards,
Muthuvel.
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DavidatK

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Joined: 22 Nov 2005
Posts: 700
Location: Troy, Michigan USA

PostPosted: Fri Dec 01, 2006 6:54 am
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muthuvel,

I know this is a late post, and you probably have your solution. But I just posted some code that uses tha RANDOM function. So if you still need it:

RANDOM FUNCTION

Dave
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muthuvel

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Joined: 29 Nov 2005
Posts: 217
Location: Canada

PostPosted: Fri Dec 01, 2006 10:30 am
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Dave,
Thanks!!!!
Though it was a late reply,Thanks for the concern.

Thanks,
Muthuvel
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