Answer / y@$w@nth

s9(9)v9(9)

Here v stands for assumed decimal point so it will occupy
only 18 bytes(9+9)...if instead of s9(9)v9(9) it is coded as
s9(9).9(9) then it is going to occupy 19 bytes ((9+9)+1)
the extra 1 byte for decimal point .........There is no way
of occupying 20 bytes storage.

Answer / jaganmohanreddy

ya we can code like this it will occupy 18 bytes only s9(9)
v9(9) v is for assumed decimal

Answer / adithya

it wil occupy 18 bytes

Answer / siddu reddy

yes, 18 bytes it will occupy. here v is assumed decimal
point,it doesn't occupy any space

Answer / sruthikarnti

thanks in advance

Answer / ramesh.p

yes, we can declare it.it is a display usage.it will occupy one byte for each character so,there are 18 characters there (9+9) .so it will occupy 18 bytes...

Answer / sruthi

thanks ....

Answer / rajagopal

For the above question the answer is 19.Because sign is
allocated automatically and in cobol user defined value
should not exceed 18.so it'll take 19bytes including sign
without giving any error.but if you give
s9(9).9(9),definitely it'll give error because it's user
defined length is 19 and with sign it'll take 20 bytes and
it'll definitely throw an error.Anyway if you give
parm.cobol=arith(extend) in compile JCL both the above cases
will work.

Answer / paddu

Yes,we can declare.s9(9)v9(9) will occupy 18 bytes. the
maximum bytes are in any comp field item is 18bytes.

Answer / karthik

NO IT CANNOT BE. THE MAXIMUM ALLOWED DIGITS IS 18 ONLY

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