Index
> Projects and Ideas > [DEAD] X to the power of...well Y |
| Author |
|
|
Madis731 28 Feb 2006, 20:56
Hi,
another program that demonstrates my bad programming habits - it was the time when I made a lot of unuseful comments like Code: ;this line increases eax by one but if you don't let it disturb, you might even find something useful out of it. What it does it take x^y like 3^4=3*3*3*3=81 etc. but it can handle large numbers like 2^131072 and still show accurate results - I might reprogram it some day to output to a file or a text-edit box 
Last edited by Madis731 on 08 Sep 2006, 10:00; edited 1 time in total |
|||||||||||
28 Feb 2006, 20:56 |
|
|
Mota 08 May 2006, 23:14
Well, I've just devised a simple recursive pow function that works exclusively for integers. It is guaranteed to work faster than your average loop, because it splits an equation in the most efficient way: down the middle.
Code: pow: ; I: eax = x, ecx = y / O: eax = x^y or ecx, ecx jnz .cont0 mov eax, 1 ret ; x^0 = 1 .cont0: cmp ecx, 1 jnz .cont1 ret ; x^1 = x .cont1: cmp ecx, 2 jnz .cont2 mul eax, eax ; x^2 = x*x ret .cont2: ; Okay, what happens is that (x^y = x^(y >> 1)^2 * x^(y & 1)), ; therefore minimizing the number of necessary operations. test ecx, 1 jz .no_one push ecx eax shr ecx, 1 call pow mul eax, eax mul eax, [esi] pop ecx ecx ret jmp .cont .no_one: push ecx shr ecx, 1 call pow ; (x^(y/2) * x^(y/2)) = (x^(y/2))^2 mul eax, eax ; so there is no need to recalculate the value. pop ecx ret This was good exercise. Another one for the code bank. |
|||
|
08 May 2006, 23:14 |
|
|
Madis731 10 May 2006, 08:41
You mean:
Code: pow: ; I: eax = x, ecx = y / O: eax = x^y or ecx, ecx jnz .cont0 mov eax, 1 ret ; x^0 = 1 .cont0: cmp ecx, 1 jnz .cont1 ret ; x^1 = x .cont1: cmp ecx, 2 jnz .cont2 mul eax ; x^2 = x*x ret .cont2: ; Okay, what happens is that (x^y = x^(y >> 1)^2 * x^(y & 1)), ; therefore minimizing the number of necessary operations. test ecx, 1 jz .no_one push ecx eax shr ecx, 1 call pow mul eax mul dword[esp] pop ecx ecx ret ; jmp .cont .no_one: push ecx shr ecx, 1 call pow ; (x^(y/2) * x^(y/2)) = (x^(y/2))^2 mul eax ; so there is no need to recalculate the value. pop ecx ret |
|||
|
10 May 2006, 08:41 |
|
|
Mota 11 May 2006, 22:12
No, I mean exactly what I wrote. What you wrote won't even compile.
Anyway, peace out. |
|||
|
11 May 2006, 22:12 |
|
|
donkey7 16 May 2006, 17:19
slightly shorter version:
Code: format pe gui
power equ 7
base equ 5
mov ecx,power
mov eax,base
mov ebx,1
.loop:
shr ecx,1
jnc .skip
imul ebx,eax
.skip:
imul eax,eax
or ecx,ecx
jnz .loop
int3 result in ebx. should work :) btw: algorithm taken from clr's 'introduction to algorithms'. interesting book! _________________ Keep coding! |
|||
|
16 May 2006, 17:19 |
|
|
vid 08 Sep 2006, 08:26
hmmm... is this really a project?
|
|||
|
08 Sep 2006, 08:26 |
|
|
Madis731 08 Sep 2006, 09:59
Well, no
a demonstration of some ideas at most... |
|||
|
08 Sep 2006, 09:59 |
|
< Last Thread | Next Thread > |
Forum Rules:
|