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muthuvel
Active User
Joined: 29 Nov 2005 Posts: 217 Location: Chennai




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
Senior Member
Joined: 28 Mar 2005 Posts: 1452 Location: Chicago, IL




The random function takes a nonnegative integer argument and returns a decimal number. The integer argument is optional. It is the seed for a pseudorandom mathematical process, that determines the functionâ€™s first returned number. In computer languages, all random processes are in fact long, repeating series of numbers that appear to be random. A wellchosen 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
Active User
Joined: 29 Nov 2005 Posts: 217 Location: Chennai




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>< 
 +(argument1)+ 
 
++
argument1
If argument1 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 argument1, a new sequence of
pseudorandom numbers is started.
If the first reference to this function in the run unit does not specify
argument1, the seed value used will be zero.
In each case, subsequent references without specifying argument1 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
Active Member
Joined: 22 Nov 2005 Posts: 700 Location: Troy, Michigan USA




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
Active User
Joined: 29 Nov 2005 Posts: 217 Location: Chennai




Dave,
Thanks!!!!
Though it was a late reply,Thanks for the concern.
Thanks,
Muthuvel 

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