How is a physical address generated in 8086?

Question

In the 8086 architecture, the memory space is 1 MiB in size and divided into logical segments of up to 64 KiB each.

i.e. it has 20 address lines thus the following method is used:

That the data segment register is shifted left 4 bits then added to the offset register

My question is: How we do the shift operation although all the registers are only 16 bits

Answer 1

Address translation is done internally by a special unit without using the registers available to user code to store intermediate results - it just fetches 16-bit values and does the translation inside - it is not reflected anywhere where the user code could observe it.