What are Segments and how can they be addressed in 8086 mode?

Question

Ever since I started with 8086 Assembly Language programming, I have been hammering my mind about these Segments and Segment registers. The problem I am facing is that I can't have a visual image of what segments are in my mind and therefore the concepts are not clear to me.

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• Can anyone help me understand the concept relating it to a real world scenario? Also I have the following questions:

Question 1:

As far as I have understood, In 16-bit real mode with 20 address line enabled, we could divide the physical memory into 16 segments with 64KiB each. The first segment starts at 0x00000. What will be the starting address of next segment. Will it be by adding 0x10000 (65536 = 64KiB)?

Question 2:

This question is a bit odd to ask here but still SO is my only option. Suppose if I am given with an Offset address of 0x6000, How can I find the segment to which it belongs in order to address it.

Thanks

Answer 1

...we could divide the physical memory into 16 segments with 64KiB each.

True, but more exact would be to phrase this as "16 non-overlapping segments" since there's also the possibility to divide the memory into 65536 overlapping segments.

When the A20 line is enabled, we have more than 1MB to play with. (1048576+65536-16) When setting the relevant segment register to 0xFFFF, we can gain access to the memory between 0x0FFFF0 and 0x10FFEF.

The main features of both kinds of segments are:

1. Non-overlapping segments
   • Contain 65536 bytes.
   • Are 65536 bytes apart in memory.
   • This is the way us people often conveniently view memory. It enables us to say that we've put
     • the graphics window in the A-segment (0xA0000-0xAFFFF)
     • the text video window in the B-segment (0xB0000-0xBFFFF)
     • the BIOS in the F-segment (0xF0000-0xFFFFF)

2. Overlapping segments

   • Contain 65536 bytes.

   • Are 16 bytes apart in memory.

     Sometimes you'll see people refer to a 16-byte chunk of memory as a segment but obviously this is wrong. There is however a widely used name for such an amount of memory : "paragraph".

   • This is the way the CPU (in the real address mode) sees memory.
     The processor calculates the linear address using next steps:
     • First is calculated the offset address from the operands of the instruction. The result is truncated to fit in 16 bits (64KB wraparound).
     • Next is added the product of SegmentRegister * 16
       If the A20 line is inactive the result is truncated to fit in 20 bits (1MB wraparound).
       If the A20 line is active the result is used as is and thus no 1MB wraparound occurs.

Suppose if I am given with an Offset address of 0x6000, How can I find the segment to which it belongs in order to address it.

Here again the problem lies in the phrasing!

If by "an Offset address of 0x6000" you mean an offset like the one we normally use in the real address mode programming then the question cannot be answered since there is such an offset 0x6000 in every segment that exists!

If on the other hand the wording "an Offset address of 0x6000" actually refers to the linear address 0x6000 then there are a lot of solutions for the segment register:

segment:offset
--------------
   0000:6000
   0001:5FF0
   0002:5FE0
   0003:5FD0
   ...
   05FD:0030
   05FE:0020
   05FF:0010
   0600:0000

As you can see there are 0x0601 possible segment register settings to get to linear address 0x6000.
The above applies to when the A20 line is indeed enabled. If A20 was inactive then the linear address 0x6000 (just like any other linear address from 0 to 1MB-1) can be reached in precisely 0x1000 (4096) ways:

segment:offset
--------------
   F601:FFF0
   F602:FFE0
   F603:FFD0
   ...
   FFFD:6030
   FFFE:6020
   FFFF:6010
   0000:6000
   0001:5FF0
   0002:5FE0
   0003:5FD0
   ...
   05FD:0030
   05FE:0020
   05FF:0010
   0600:0000