x86 Assembly from my understanding

Soooo this article (or maybe even a series of articles, who knows ?) will be about x86 assembly, or rather, what I understood from it and my road from the bottom-up hopefully reaching a good level of understanding

Memory :

Memory is a sequence of octets (Aka 8bits) that each have a unique integer assigned to them called The Effective Address (EA), in this particular CPU Architecture (the i8086), the octet is designated by a couple (A segment number, and the offset in the segment)

The Segment is a set of 64 consecutive Koctets (1 Koctet = 1024 octets).
And the offset is to specify the particular octet in that segment.

The offset and segment are encoded in 16bits, so they take a value between 0 and 65535

Important :

The relation between the Effective Address and the Segment & Offset is as follow :

Effective address = 16 x segment + offset keep in mind that this equation is encoded in decimal, which will change soon as we use Hexadecimal for convention reasons.

Example :

Let the Physical address (Or Effective Address, these two terms are enterchangeable) 12345h (the h refers to Hexadecimal, which can also be written like this 0x12345), the register DS = 1230h and the register SI = 0045h, the CPU calculates the physical address by multiplying the content of the segment register DS by 10h (or 16) and adding the content of the register SI. so we get : 1230h x 10h + 45h = 12345h

Now if you are a clever one ( I know you are, since you are reading this <3 ) you may say that the physical address 12345h can be written in more than one way….and you are right, more precisely : 2¹² = 4096 different ways !!!

Registers

The 8086 CPU has 14 registers of 16bits of size. From the POV of the user, the 8086 has 3 groups of 4 registers of 16bits. One state register of 9bits and a counting program of 16bits inaccessible to the user (whatever this means).

General Registers

General registers contribute to arithmetic’s and logic and addressing too.

Each half-register is accessible as a register of 8bits, therefor making the 8086 backwards compatible with the 8080 (which had 8bit registers)

Now here are the Registers we can find in this section:

AX: This is the accumulator. It is of 16 bits and is divided into two 8-bit registers AH and AL to also perform 8-bit instructions. It is generally used for arithmetical and logical instructions but in 8086 microprocessor it is not mandatory to have an accumulator as the destination operand. Example:

ADD AX, AX ;(AX = AX + AX)

BX: This is the base register. It is of 16 bits and is divided into two 8-bit registers BH and BL to also perform 8-bit instructions. It is used to store the value of the offset. Example:

MOV BL, [500] ;(BL = 500H)

CX: This is the counter register. It is of 16 bits and is divided into two 8-bit registers CH and CL to also perform 8-bit instructions. It is used in looping and rotation. Example:

MOV CX, 0005
LOOP

DX: This is the data register. It is of 16 bits and is divided into two 8-bit registers DH and DL to also perform 8-bit instructions. It is used in the multiplication and input/output port addressing. Example:

MUL BX (DX, AX = AX * BX)

Addressing and registers…again

I realized what I wrote here before was almost gibberish, sooo here we go again I guess ?

Well lets take a step back to the notion of effective addresses VS relative ones.

Effective = 10h x Segment + Offset . Part1

When trying to access a specific memory space, we use this annotation [Segment:Offset], so for example, and assuming DS = 0100h. We want to write the value 0x0005 to the memory space defined by the physical address 1234h, what do we do ?

Answer :
```
MOV [DS:0234h], 0x0005
```
Why ? Let’s break it down :

We Already know that Effective = 10h x Segment + Offset, So here we have : 1234h = 10h x DS + Offset, we already know that DS = 0100h, we end up with this simple equation 1234h = 1000h + Offset, therefor the Offset is 0234h

Simple, right ?, now for another example

Another example :

What if we now have this instruction ?

    MOV [0234h], 0x0005

What does it do ? You might or might not be surprised that it does the exact same thing as the other snipped of code, why though ? Because apparently and for some odd reason I don’t know, the compiler Implicitly assumes that the segment used is the DS one. So if you don’t specify a register( we will get to this later ), or a segment. Then the offset is considered an offset with a DS segment.

Segment + Register <3

Consider DS = 0100h and BX = BP = 0234h and this code snippet:

    MOV [BX], 0x0005 ; NOTE : ITS NOT THE SAME AS MOV BX, 0x0005. Refer to earlier paragraphs

Well you guessed it right, it also does the same thing, but now consider this :

    MOV [BP], 0x0005

If you answered that its the same one, you are wrong. And this is because the segment used changes according to the offset as I said before in an implicit way. Here is the explicit equivalent of the two commands above:

    MOV [DS:BX], 0x0005
    MOV [SS:BP], 0x0005

The General rule of thumb is as follows :

If the offset is : DI SI or BX, the Segment used is DS.
If its BP or SP, then the segment is SS.

Note
The values of the registers CS DS and SS are automatically initialized by the OS when launching the program. So these segments are implicit. AKA : If we want to access a specific data in memory, we just need to specify its offset. Also you can’t write directly into the DS or CS segment registers, so something like
```
MOV DS, 0x0005 ; Is INVALID
MOV DS, AX ; This one is VALID
```