Why does 8086 segment addressing not overflow?

Question

I know that the 8086 addressing mode is a left shift of the segment register by four bits then add an offset. But will the information of significant bit not be lost after shifting the 16-bit register?

For example, the segment register stores 0x1000, will it not become 0x0000 after shifting four bits to the left?

Answer 1

Address calculation happens with 20-bit math. Or more on later CPUs. Not the same width as the inputs. The CPU doesn't shift inside the segment reg!

In CPUs that don't support other modes (like protected mode where the base comes from the GDT), I assume the selected segment register is just hard-wired to bits [20:4] of an adder in the AGU, so the "shift" is just built-in to the wiring. Not like when you run a shift instruction and the result goes back into the destination.

(Actual 8086 didn't have an AGU separate from its ALU, so presumably that's what happens when using that block of logic to do address math instead of 16-bit normal integer math. It also supports a mul instruction which produces a 32-bit product, that has to get split across DX:AX when it's done, but internally the ALU has to shift and add numbers across 16-bit boundaries for multiply to work.)