hexadecimal value of opcodes

Question

I created a very simple assembly program that prints the letter 'a' in DOS. I opened it in a hex editor and the result was this:

Assembly code:

mov ah, 2 
mov dx, 'a' 
int 21h 

Hex code

B4 02 B2 61 CD 21

I wanted to understand how it was generated! Like, I do not know if I'm right, but I realized that:

B4 = mov ah 
02 = 2 
B2 = mov dx 
61 = 'a' 
CD = int 
21h = 21

The 02, 61 and 21 I understood what turned but and B4, B2 and CD?

Solution 1

Here's a nice reference: http://ref.x86asm.net/coder32.html

As you can see:

• CD is the opcode for int
• B0+reg is the opcode for mov reg, imm8, where reg is the destination register and as you can see from this table, ah = 100b and dx = 010b

Solution 2

Are Assembly x86 instructions:

• B4: mov ah mean move in the register ah
• B2: mov dx mean move in the register dx
• CD: int means software interrupt

I recommend you read this guide assembly x86 http://www.cs.virginia.edu/~evans/cs216/guides/x86.html

Comments

[pasztorpisti]

The official docs: intel.com/content/www/us/en/processors/… If you are interested only in "instruction opcodes", the bytes into which different instructions are encoded then on the previously linked site start with this document: "Manual Volume 2A: Instruction Set Reference, A-M" The site linked by an answer below (ref.x86asm.net/coder32.html) is a very good summary/overview of the intel PDFs but read the intel docs to learn the exact behavior of instructions.

[pasztorpisti]

BTW, if you are thinking about playing around with assembly and maybe with disassembling/reverse engineering then try the best disassembler/debugger of all times, it has a free version (5.0) that knows much less then the newer versions but even this old free version can kick the ass of any other solution: hex-rays.com/products/ida/support/download_freeware.shtml It can come handy in analyzing stuff.

[Sourav Kannantha B]

since there are 16 general purpose registers, it would require 4 bits to encode them.. isn't it?

[Sourav Kannantha B]

in the reference you provided, they have given opcodes and equivalent hex values. But I am not understanding how to combine them to obtain valid instruction. For eg, 0x817 corresponds to cmp, and 0x5 corresponds to reg ebp. But with this information, how can I encode cmp dword [ebp-4] 2. I have searching through internet for this for around 2 hrs. Help me out!!