How many bytes will be allocated for the following record
description
entries?
01 REC-A.
05 A PIC S9(4).
05 B PIC XXXBXXX.
05 C PIC ____9.99.
05 D PIC S9(5) COMP-3.
05 E PIC 9(3) COMP.
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Answer Posted / vijay
05 A PIC S9(4) -> 5 (BY CONSIDERING ONE SIGN BIT)
05 B PIC XXXBXXX -> 7 ( AS YOU ALL KNOWS ABOUT THIS)
05 C PIC ____9.99 -> 4 (SPACE AND 9V99=9(3))I THINK
05 D PIC S9(5) COMP-3 -> 3 ((5/2)+0.5)FOR ODD NUM WE HAVE
CONSIDER ONE 1/2 BYTE ONLY=0.5)
05 E PIC 9(3) COMP -> 2 (9(1) TO 9(4)=HALFWORD=2BYTES)
(9(5) TO 9(9)= WORD= 4 BYTES)
(9(10) TO 9(18)= FULLWORD=8BYTES)
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