Lace

Gave it a look-see, don't see anything too bad. Suggestions though, in the bullet segment you said <framerects>, you should use my calculation method instead, saves tons of space. I uploaded it.
Also, you IMUL ECX by -1 in C of the Bullet, I think NEG ECX does the same thing but faster does it not? Or am I mistaken.

Haven't checked yet D:
But I'm glad I won't have to find it myself :]
Thanks bunches

I get a different answer, I get that the area of the rectangle is:
(2/a) * (b^2 + 4a^2 - 4ac)
or, if you prefer:
2b^2/a + 8a - 8c

If a is positive, and the answer you get is positive, or if they're both negative, then you're all right (and you just take the absolute value if it's negative). If they have a different sign, however, it means that the quadratic and its derivitive do not intersect, so you can't make a rectangle. If the answer you get is zero, then that means that the quadratic and its derivative intersect at a single point (i.e. are tangent to each other). If a is zero, of course you get no answer because you divide by zero.

That seems weird, let's see:

f(x) = ax^2 + bx + c
f'(x) = 2ax + b
ax^2 + (b-2a)x + (c-b) = 0
x = ((2a-b) +/- sqrt((b-2a)^2 + 4a(b-c)))/(2a)
x = ((2a-b) +/- sqrt(b^2 + 4a^2 - 4ac))/(2a)
y = (2a-b) +/- sqrt(b^2 + 4a^2 - 4ac) + b

And I don't have time to figure out the rest, but it doesn't work if b^2 + 4a^2 - 4ac is less than zero, cause then they don't intersect and you don't have a rectangle :eek:. Yours doesn't account for that

Amen to that! Thanks for the support ;D

I'm the same way, for the most part, I really don't like to rely on people to do something for me, because:
-it won't be mine, then
-it might not be exactly the way i intended it

It's definitely a challenge for me, but one that I'm eager to take up!

I will eventually, but for now, I'd really like to learn how to make a mod that I'd wanted to do ever since I played my first CS mod 8D

Been working on my mod more than I have that, It's kinda on hold D:
Mod's going fairly well though as far as spriting goes, so I'm proud with that

I added a bunch of stuff, filled out some tables, added some tables, functions, etc.
I'll next add some frequently used offsets in the manner you did then be done for a little while, since I have another exam tomorrow.

This looks pretty cool. Is there an organizational reason you have different bits like [49e654] in individual notes? Like how should I set things up.

Sounds cool. I don't know what evernote is but if it's got tags then that must be good. Good luck, I'd like to see it finished.

No problem.
And, encryption is silly, and I don't even rightly know how it works. Best to just leave it be..

There's a good chance that when you call 41e136, whatever's in ECX isn't going to stay there. Make a temp variable or put it into an [eax+n]

Oh, I think I see it. Functions return to EAX, not ECX, is that not it

Umm... it looks like it should do something. Well, you can check what's happening by setting a breakpoint at 41e110 and pressing F8 once it catches to follow the thing step by step, that's usually my solution. It could be a misdirected JE/JNE or something like that.

I figured it might be possible to have a data segment where you would put the entire ASM code for the game, and then apply some sort of encryption like maybe adding 7 to every byte less than 99. When the game starts, it would "Decrypt" the data by undoing whatever it was that was done, but beforehand the game can do a check to see if there is a debugger running and stop execution before it is decrypted if there is.

Because, ASM code can be put like where the old map data was and still run [basically]
But, really that's a pretty far-fetched scheme, and personally I don't think it would be worth it.

interesting idea. I tried it just now, the machine wasn't able to recognize it as a windows executable though so it couldn't play. Thing is, adding those extra bytes [or, conversely subtracting some] would pretty much change every non-short jump and every call, wouldn't it?

There ya go!

(in your "perfect world", multiplying something by zero would be like dividing it by two and squaring it O_o).

Ready to take another stab at that 15 question quiz I showed you earlier?

On the contrary, it's much cooler that (x-y)(x+y) is a quadratic function with respect to both x AND y. If it were like your way, it would only be with respect to x, but not y.