CODEWITHSUNDEEP
assemblyx86cpu-registers
Andrei Vlad
Andrei Vlad

Reputation: 11

Basic stuff assembler (TASM)

I'm new to assembly language and I would like to ask you the following:

    mov ax, y
imul z; dx:ax = y*z
mov bx, dx
mov cx, ax ; bx:cx = y*z
mov ax, x
cwd ; dx:ax = x
sub ax, cx
sbb dx, bx ; dx:ax = x-y*z

Why do I have to use the last code line. What am I trying to do is to calculate x-y*z...

Thank you in advance

Upvotes: 0

Views: 284

Answers (1)

Codie CodeMonkey
Codie CodeMonkey

Reputation: 7946

If I recall my 80386 assembly (yes, it's been that long), the last line is a subtract with borrow. This is the same as when you do subtraction by hand: if you subtract 16 from 24, for example, you subtract 6 from 4 first. But to do that you BORROW a 1 from the tens column and get 12-6=6 for the low order digit. When you do the high order digit, you have to remember that you borrowed and take that into account: 10-10 =0, so the answer is 6.

When you sub ax,cx, the borrow flag is set if borrowing was necessary, when you sbb dx, bx, you adjust for the previous borrow.

Upvotes: 2

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