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Messages - jnesselr
Pages: 1 ... 52 53 [54] 55 56 ... 165
796
« on: February 27, 2011, 02:35:00 pm »
tiDE cannot compile KnightOS at the moment, and I'm way to busy to fix it.
mmk, that's fine. Good luck with your "talk". I wish you well!
797
« on: February 27, 2011, 02:15:35 pm »
That looks amazing. I almost wanted to downrate so you would stay at 256 respect, but alas, I couldn't do it. * King Graphmastur notices that Builderboy has 4097 posts, and goes to delete one...
798
« on: February 27, 2011, 02:13:48 pm »
Thanks for the encouragement. You can probably expect the kernel to be released two weeks after I'm ungrounded, and the developer preview another two weeks after that.
But since you're grounded for infinity more years... But seriously, when do I need to get the filesystem to work on USB? I'm sure you kinda have your timeline, and I just wanted to know if you were going to have a way to send programs, or what. Also, is tiDE ready enough to compile KOS for us? I'm going to check out a book from my library to study about mass storage and USB for this.
799
« on: February 27, 2011, 02:06:33 pm »
I'll take some pudding, only because the Japanese paku paku is essentially nom nom. (not exactly, but hey, details...)
800
« on: February 26, 2011, 09:57:07 pm »
Actually, you also need this to work over gCn. ;-)
This is actually awesome. I love the way you do grayscale.
801
« on: February 26, 2011, 09:53:40 pm »
I need to be able to prove it for all numbers even if I prove it for the first trillion there are still an infinite more. Is there any way you could prove this in a different way?
Sure, easily. Let's try it with 1 and 2. (1)(2)=2, correct? Okay, how about 2 and 3? (2)(3)=6. So, we could say (N)(N+1)=N 2+N. Now, is this true? 1 2+1=2. 2 2+2=6. Actually, because this works for (N)(N+1), we can try and replace N with N+1. So that gives (N+1)(N+1+1) or (N+1)(N+2). Whoa, that looks familiar. (N+1)(N+2)=N 2+3N+2. And, this works for all N, since we tested it for N and N+1. This is because every time we do N+1 as N, it still works. Now then, let's try mixing it up a little bit. How about (N+X)(N+1). Whoa, that's interesting. (N+X)(N+1) = N 2+NX+N+X. Now, since X is just some number as well, and we already proved (N+1)(N+2) = N 2+3N+2, and that it can work for any N, obviously this must work as well. The same could be said about (N+1)(N+Y) = N 2+N+NY+Y. Now, for the interesting part. (N+X)(N+Y)=N 2+XN+YN+XY for any N (as we proved), any X (as we proved), and any Y (as we proved). Wait, we just proved the entire thing. In fact, to go backwards, check this out: Given N 2+XN+YN+XY, if we take N out of the first two terms, we get N(N+X)+YN+XY. And Y out of the last two terms gives N(N+X)+Y(N+X). And since you have (N+X) for both the N and the Y, you can do (N+Y)(N+X). And you can do that, because of being able to distribute it. EDIT: If you liked it, a thumbs up would be nice.  * King Graphmastur hopes this is useful either way
802
« on: February 26, 2011, 09:44:27 pm »
That's exactly what I'm talking about.
Then don't just respond with yes... meh. Anyway, now that that's settled, and we all have good coding practice.... What are you going to choose for undos? You can't have infinite undos the way you are talking about. Not unless you save the state that happened right before the last update for every single update. (Which isn't possible) I'd go with my previous idea, but then again, I'm biased, and am not entirely sure how hard it would be to code.
803
« on: February 26, 2011, 08:31:15 pm »
Yes.
slightly spamish, but okay. But wait, was that yes to me, or.... And it's good to program like that, but I don't make that my absolute top priority.
804
« on: February 26, 2011, 08:28:20 pm »
That looks interesting. Don't forget Death Note! The second one is somewhat creepy. 1st one: 2nd one: Can't forget naruto!:
805
« on: February 26, 2011, 07:50:36 pm »
Yes.
yes, to whom exactly? TravisE?
806
« on: February 25, 2011, 10:56:42 pm »
You would have to change some variables. Essentially check from the basic_start and basic_end (There's also one for basic_pc or something like that) for your label, and set basic_pc to it.
EDIT: That should work, if it doesn't, don't say I didn't warn you. btw, I'm warning you now. ;-) Not that I think it'll do anything bad.
807
« on: February 25, 2011, 10:54:26 pm »
Okay, here is the Axe starting code for an APP:
CD8B40 ;jumps to the program code
EF364C ;gets ready to exit
CD50002740 ;exits
I believe that last one can be a simple EF2740 because that bcall automatically exits (meaning you don't need to use bjump)
No, you have to use bjump here. In fact, if you didn't, it would load stuff on the stack that isn't taken off, which is bad.
808
« on: February 25, 2011, 10:51:05 pm »
You could actually optimize it, and kinda cheat it a little bit. If you have to go back one step, you have the picture, because you stored it right before the new step. But if you have to go back any farther, or hit any other redraws, you have the last element be run right after you store the picture.
809
« on: February 25, 2011, 10:44:22 pm »
I like how it essentially randomly chooses stuff, so you can give criteria even though you don't fully know where it's gonna go.
810
« on: February 25, 2011, 10:35:58 pm »
mods can be really helpful sometimes. 
And a pain if you're trying to crack security systems... But I guess useful if you wanted them secure.
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