idiv instruction in assembly 8086

Asked Dec 22 '14 at 01:33

Active Dec 22 '14 at 01:33

Viewed 983 times

I'm trying to find the value of AX.

AX = (DX AX) / operand DX = remainder (modulus).

All the values are in hexademical

mov CX, 80A2
mov AX, DD38
mov DX, 49
idiv CX
add AX, DX

AX-?

What am I doing wrong? This is what I do:

AX = 0049 DD38 / 80A2 = 93
DX = 0049 DD38 mod 80A2 = 32
AX = 93 + 32 = C5

The correct answer is 3A4C

asked Dec 22 '14 at 01:33

user2879175

3

Since you are using idiv (and not div), 80A2 is interpreted as a negative number... – Matteo Italia Dec 22 '14 at 01:41
Tried 0 - 80A2 = 7F5E, but didnt help. Any more clues ? :) Cant find anything simple with Google. – user2879175 Dec 22 '14 at 01:59
2

Look for "2's complement representation". 80A2 is not interpreted as 32930, but as -32606. Also the output is going to be in 2's complement. – Matteo Italia Dec 22 '14 at 02:01
So it goes: 80A2 -> 7F5E, then 49DD48 mod 7F5E = 94 and div = 3AE0. And 3AE0 - 94 = 3A4C. Thanks :) – user2879175 Dec 22 '14 at 02:30

0 Answers0