CODEWITHSUNDEEP
assemblyx86hexdos
user3500017
user3500017

Reputation: 197

hexadecimal value of opcodes

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?

Upvotes: 6

Views: 31237

Answers (2)

pNre
pNre

Reputation: 5376

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

Upvotes: 7

invictus1306
invictus1306

Reputation: 587

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

Upvotes: 2

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