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

Solution 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

Solution 2:

In this answer, I am only giving an explanation for real mode. In protected mode, segementation is a bit more complicated and as you're probably never going to write a segmented protected mode program, I'm not going to explain this.

Segments are very simple actually. The 8086 CPU has four segment registers named cs, ds, es, and ss. when you access memory, the CPU computes the physical address like this:

physical_address = segment * 16 + effective_address

where effective_address is the address indicated by the memory operand and segment is the content of the segment register for this memory access. By default, cs is used when the CPU fetches code, ss is used for stack pushes and pops as well as memory operands with bp as the base register, es is used for certain special instructions and ds is used everywhere else. The segment register can be overridden using a segment prefix.

What does that mean in practice? The 8086 has 16 bit registers, so using a register to store an address allows us to address up to 65536 bytes of RAM. The idea behind using segment registers is that we can store additional bits of the address in a segment, allowing the programmer to address a bit more than 2²⁰ = 1048576 bytes = 1 MiB of RAM. This RAM is sliced into 65536 overlapping segments of 65536 bytes each, where each segment is one value you can load into a segment register.

Each of these segments starts at an address that is a multiple of 16 as you can see in the address computation logic above. You can tile the entire 1 MiB physical address space with 16 non-overlapping segments (as you explained in your question) values 0x0000, 0x1000, ..., 0xf000 but you can use any segment selector you like as well.