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The 8086/8088 is a 16 bit processor running on a 16 bit (8086) or 8 bit (8088) bus with a 20 bit address bus. In order to obtain the extra 4 bits of addressibility, Intel designed segment registers that are effectively multiplied by four and then added to the 16 bit offset address generated by the instruction. This yields 64K segments of 64KB each, although they overlap each other at a distance of 16 bytes.

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Memory segmentation in the 8086/8088 is used to allow the processor to access more than 64kb of memory, even though it is only a 16-bit processor. Each segment register allows access to one of 64k 64kb segments, each overlapping by 16 bytes, with the total addressibility being 1mb.

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You really should forget about it; the segment-offset memory model of 8086 has been nothing but a pain since it was invented in 1978 (32 years before!)

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15y ago
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Q: Why you Use Memory Segmentation In 8086 Microprocessor?
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When does segmentation fault occur for a prog?

You either reference memory that is non existent, or you attempt to modify memory that is read only. This is usually a result of failure to properly initialize or use pointers or arrays.


What is use of offset address in 8086 mp?

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What is the working of 8087 NDP microprocessor?

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Maybe you mean the prefetch queue?


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