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drajasekharreddy · Beginner Joined: 08 Apr 2004 Posts: 16 Topics: 14

Hi,

my question is s9(9) comp will occupy 4 bytes. and S9(10) comp will occupy 8 bytes. this everybody knows. but my doubt is why s9(9) comp is taking 4 bytes where as s9(10) comp is taking 8 bytes, even thou the difference is 1 number between s9(9) comp and s9(10) comp. can any body know the reason. r we have to remebeber like this .

s9(1-4) comp = 2 bytes
s9(5-9) comp = 4 bytes
s9(10-18 ) comp = 8 bytes.

Thanx
Raja

kolusu · Posted: Mon Jun 21, 2004 6:30 am Post subject:

drajasekharreddy,

The storage of the binary items is in halfword ,fullword and doubleword which is the reason for 2, 4 , 8 bytes

1 through 4 digits occupy 2 bytes (halfword) and 5 through 9 digits occupy 4 bytes (fullword) and 10 through 18 occupy 8 bytes (doubleword)

Hope this helps...

Cheers

Kolusu
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