操作码的十六进制值

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 ?

Answer 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

Answer 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