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Doubt regarding 8086 RealMode Address Calculation

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      03-18-2005, 11:49 AM
In 8086 real mode addressing the 16 bit segment register is added with
16 bit offset and 20 bit address is generated.
Case 1:
Segment Reg: B230
Offset Reg: 2052

Now the resulting 20 bit address is generated as
Segment reg+ B230+
offset 02052
Address = B4352
The resulting address generated is accessed.

Case 2
Now my doubt is what if Segment Register: FFFF and offset register:
FFFF value
in it.If we calculate the Resulting 20 bit address it will cause an
overflow Right!!!!
Can somebody clarify how 20 bit address is generated in this case?????

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