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My teacher has given me the question to differentiate the maximum memory space of 1MB and 4GB microprocessor. Does anyone know how to answer this question apart from size mentioned difference ? https://i.stack.imgur.com/Q4Ih7.png

Peter Cordes
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kaito kumon
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  • Real-mode segments overlap, they're only 16 bytes apart but are 64KiB long. – Peter Cordes Nov 16 '22 at 08:21
  • "the difference in memory space between 1MB and 4GB in 8086/88" is illogical, because the 8086/88 never supported 4GB. It is like "the difference in wheels between 4-wheeled and 16-wheeled Tesla model 3": there is no such thing as a 16-wheeled Tesla model 3. – Mike Nakis Nov 16 '22 at 10:16

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A 32-bit microprocessor can address up to 4 GB of memory, because its registers can contain an address that is 32 bits in size. (A 32-bit number ranges from 0 to 4,294,967,295‬). Each of those values can represent a unique memory location.

The 16-bit 8086, on the other hand, has 16-bit registers which only range from 0 to 65,535. However, the 8086 has a trick up its sleeve- it can use memory segments to increase this range up to one megabyte (20 bits). There are segment registers whose values are automatically bit-shifted left by 4 then added to the regular registers to form the final address.

For example, let's look at video mode 13h on the 8086. This is the 256-color VGA standard with a resolution of 320x200 pixels. Each pixel is represented by a single byte and the desired color is stored in that byte. The video memory is located at address 0xA0000, but since this value is greater than 16 bits, typically the programmer will load 0xA000 into a segment register like ds or es, then load 0000 into si or di. Once that is done, the program can read from [ds:si] and write to [es:di] to access the video memory. It's important to keep in mind that with this memory addressing scheme, not all combinations of segment and offset represent a unique memory location. Having es = A100/di = 0000 is the same as es=A000/di=1000.

puppydrum64
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