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I learnt that the physical address is calculated by shifting the segment address (16-bit) left 4 times and adding it with the 16-bit offset address. The memory in the 8086 architecture is 1M.
My question is if the segment register and the offset value both are FFFFH and FFFFH then the result would be more than FFFFH i.e., more than 1M.
haw is it actually calculated…??
It does modular arithmetic, dropping any carries. So for a segment of FFFF and offset of FFFF, you compute FFFF0 + FFFF = 10FFEF but it “drops” the initial 1, leaving a real answer of 0FFEF.