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Doubt regarding 8086 RealMode Address Calculation

 
 
mark_india
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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.
Eg:
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?????


Regds
PS
 
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