by whale » Mon Jan 21, 2008 4:39 pm
This also helps .
FIXED DEC(5,2) takes
5*(1/2 byte) + 1/2 byte for sign = 3 bytes
FIXED DEC(4,2) or FIXED DEC(4,0) (as is the case here).
4*(1/2 byte) + 1/2 byte for sign = 2 and half bytes. But the remaining 1/2 byte is wasted. So it too occupies 3 bytes.
FIXED DEC in PL1 is equal to COMP-3 in Cobol
eg: FIXED DEC(5,1) = 9(4)v9(1) comp-3
Regards ,
whale
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