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If Core Segment (CS) =2000 and Instruction Pointer = FFFE. What is the next address generated for fetching the instruction by 8086 processor?

It cannot be 30000 as far as I know because the last address of CS is 2FFFF. So I think it will go back and finally reach 20000 ie the starting address. Please confirm.

Evil_Transistor
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  • Did you try computing it? What do you think it is? Note: address is `(CS << 4) + IP`. Really your only question then should be: when the 8086 increments the instruction pointer (IP), does it update the CS register if it overflows or does it leave it alone? If you can answer that, then you can answer your original question. Additional useful information: [What is the purpose of CS and IP registers in Intel 8086 assembly?](https://stackoverflow.com/questions/17777146/what-is-the-purpose-of-cs-and-ip-registers-in-intel-8086-assembly). – lurker Jan 24 '18 at 17:06
  • The highest accessible via CS:IP is FFFFF not 2FFFF. It doesn't really make sense to say, *the last address of `CS` is `2FFFF`*. – lurker Jan 24 '18 at 17:25

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