Arithmetic and Logic

Group I: add, sub, cmp, and, test, or, xor

The following instructions can affect the flags CF, ZF, SF, OF, PF, and AF.

Assume X = reg | mem and Y = reg | mem | immed

add X, Y ; X = X + Y

sub X, Y ; X = X – Y

cmp X, Y ; X – Y

and X, Y ; X = X & Y

test X, Y ; X & Y

or X, Y ; X = X | Y

xor X, Y ; X = X ^ Y

The mul instructions multiples unsigned ints. Assume X8 is an 8-bit register or memory location, etc. Then:

mul X8 ; ax = al * X8

mul X16 ; dx:ax = ax * X16

mul X32 ; edx:eax = eax * X32

There are equivalents of these for imul, the instruction of multiplying signed ints. In addition, there are other formats such as:

imul R32, X32, I32 ; R32 = X32 * I32

Where R32 is a 32-bit register, X32 is a 32 bit register or memory reference, and I32 is a 32 bit immediate.

The mul and imul instructions can affect the CF and OF flags.

The instructions for dividing two unsigned ints are:

div X8 ; al = ax / X8

div X16 ; ax = dx:ax / X16

div X32 ; eax = edx:eax / X32

The formats are the same for idiv, the instruction for dividing signed ints.

These instructions don't affect the flags.

Assume X= reg | mem

inc X ; X = X + 1 affects ZF, SF, OF, PF, AF

dec X ; X = X – 1 affects ZF, SF, OF, PF, AF

not X ; X = ~X does not affect flags

neg X ; X = -X affects CF, ZF, OF, PF, AF

How many bricks does it take to build a staircase of height n?

Clearly the answer is 1 + 2 + 3 + ... + n. This is called the nth triangle numbers (because the staircase is shaped like a right triangle).

One trick is to rearrange the sum as follows:

(1 + n) + (2 + n – 1) + (3 + n – 2) + ... = n * (n + 1) / 2

Here's the solution: