What will be the result of these instructions.
Assume that BYTE1 is defined as DB 05.
BEFORE
a.) MOV CX, 25H ;CX = 0000H
b.) MOV CL, 0 ;CX = FFFFH
c.) MOV AX, BYTE1 ;AX = 1234H
d.) ADD DL, BYTE1 ;DX = 0120H
e.) INC DX ;DX = FFFFH
f.) INC DL ;DX = FFFFH
g.) XCHG AH,AL ;AX = 1234H
h.) SUB CX,CX ;CX = 1234H
i.) XCHG CX, CX ;CX = 1234H
I got these from my cousin's notebook in his class in assembly 2 years ago, and I can't figure it out the final result.
a.) MOV CX, 25H ;CX = 0000H
CX = 0025h
b.) MOV CL, 0 ;CX = FFFFH
CX = 0FF00h
c.) MOV AX, BYTE1 ;AX = 1234H
we do not know what BYTE1 is, but you can't add a byte operand to a word register
d.) ADD DL, BYTE1 ;DX = 0120H
we do not know what BYTE1 is
e.) INC DX ;DX = FFFFH
DX = 0000
f.) INC DL ;DX = FFFFH
DX = 0FF00h
g.) XCHG AH,AL ;AX = 1234H
AX = 3412h
h.) SUB CX,CX ;CX = 1234H
CX = 0000
i.) XCHG CX, CX ;CX = 1234H
CX = 1234h
Quote from: j053f on January 22, 2011, 10:22:35 AM
I got these from my cousin's notebook in his class in assembly 2 years ago, and I can't figure it out the final result.
:cheekygreen:
Why not type them into DEBUG and
watch what happens?
To dedndave, thank you very much for your help. I hope I can still ask your help if you don't mind. God bless you.
To redskull, thank you very much for your help, I did not know that these the result of these instructions can be viewed in debug command. Now I know, thank you once again.