x86 partial register usage

Question

If I save a value, let's say 10, in 8 bit register DH and then another value, 15, in 8 bit register DL. Would that work or will they override each other since they are both in 32-bit EDX register?

mov $10, %DH
mov $15, %DL
cmp %DL, %DH

jle done

Basically I'm just confused when I'm using the 8 bit register how will it affect the 32 bit register and vice versa. Thanks.

Also, can you save the value 7 in EDX and DH and DL would still have their own values or will they now have 7?

Answer 1

DL is the least significant byte of DX, and DH is the most significant byte of DX. DX in turn is the least significant word of EDX.

So:

MOV EDX,0x12345678
; Now EDX = 0x12345678, DX = 0x5678, DH = 0x56, DL = 0x78

MOV DL,0x01
; Now EDX = 0x12345601, DX = 0x5601, DH = 0x56, DL = 0x01

MOV DH,0x99
; Now EDX = 0x12349901, DX = 0x9901, DH = 0x99, DL = 0x01

MOV DX,0x1020
; Now EDX = 0x12341020, DX = 0x1020, DH = 0x10, DL = 0x20

As you can see, you can write to DL or DH without them affecting one another (but you're still affecting DX and EDX).

---

Also, can you save the value 7 in EDX and DH and DL would still have their own values or will they now have 7?

As you can deduce from my examples above, DH would get the value 0 andDL the value 7.

Answer 2

If you address DH and DL individually, then the values are individually accessible and be kept as long as you don't perform operations which will affect other parts of the register.

For example

                       EDX------
                            DX --
                            DH DL
xor edx, edx   ; edx = 0000 00 00
mov 01h, dh    ; edx = 0000 01 00
mov 02h, dl    ; edx = 0000 01 02
sub dh, dl     ; edx = 0000 ff 01  <- Note that this doesn't
               ; overflow to the high word, because you
               ; are using only 8 bit registers
mov 80000001, eax
sub edx, eax   ; edx = 8000 FF 03   <- Here you will affect
               ; the whole register, because you address it with 32 bit.