title: Calculation of physical address in 8086
body: <p>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.</p>
  
  <h2>    FFFF0</h2>
  
  <h2>+   FFFF</h2>
  
  <h2>----------</h2>
  
  <h2> 10FFEF</h2>
  
  <p>haw is it actually calculated...??</p>