Point to remember: Even if we specify 'S' or not in the variable declaration, the sign is stored.
Suppose we move the digit 1265875 into a field 9(7). In display mode, cobol will occupy 7 bytes, ie, 7 storage positions.
If we use COMP3 in the variable declaration, it will take only 4 Bytes like below.
12

65

87

5C

Had it been a negative number, then The rightmost digit would store 'D' in place of C.
How many bytes does COMP3 take ?
To calculate the bytelength of a comp3 field,
start with the total number of digits and divide by 2 giving a result
(discarding the remainder if any), then add 1 to the result.
Note: For "normal" processing the maximum number of digits is 18 . If we use compiler option as the ARITH(EXTEND) , the compiler will allow us to extend it to 31 bytes
Just to share, if we use Odd number of digits for packed decimal, the processing becomes 520 % faster than what the speed would be if we use even number of digits!. (Source : Some Informative journal in comp3 in google )
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