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Topic: Graviter (Read 182836 times)
0 Members and 2 Guests are viewing this topic.
leafy
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31337 u53r (Next: 2000)
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Seizon senryakuuuu!
Re: Graviter - Axe
«
Reply #255 on:
March 10, 2011, 01:54:40 am »
Could I achieve the same effect doing something like
{L2}r/256*256-->{L2}r?
I don't quite understand how yours works.
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Builderboy
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Re: Graviter - Axe
«
Reply #256 on:
March 10, 2011, 01:57:47 am »
not quite, since that makes the variable smaller, when you want to be making it larger. adding 255 would work though.
If this: 23.156 is your Y value, you are on the 23rd pixel, and 156/256 the way down the pixel. We want to be at the very bottom, we want to be at 23.255, so we want to set the second half to 255. We can do OR 255, we can try /256*256+255, or 255->{oY}. whatever we like
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leafy
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Re: Graviter - Axe
«
Reply #257 on:
March 10, 2011, 01:59:12 am »
Ah I think I see now. I'll try this when I have time.
Is the o symbol the same one in the section with the r token? I'll try it and if it doesn't work I'll use your other method.
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Builderboy
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Re: Graviter - Axe
«
Reply #258 on:
March 10, 2011, 02:00:41 am »
yeah, its the symbol used to find the address of the Y variable, that way I can set the specific byte that corosponds to it
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leafy
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Re: Graviter - Axe
«
Reply #259 on:
March 10, 2011, 02:06:47 am »
Wait so how does the Or work? I thought it's a boolean, so it wouldn't be adding anything to y? Or am I wrong?
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Builderboy
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Re: Graviter - Axe
«
Reply #260 on:
March 10, 2011, 02:09:42 am »
It is a boolean, but it works on each individual bit of the number
Since your number is 10101001.10010010 , and since we want the second half to be 11111111, we just OR your number with 11111111 (255), which sets all of the bits to 11111111
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leafy
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Re: Graviter - Axe
«
Reply #261 on:
March 10, 2011, 02:12:23 am »
It is starting to make some vague sense in my mind. Sorry, I'm just used to thinking of boolean as 0 and 1 xD
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Builderboy
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Re: Graviter - Axe
«
Reply #262 on:
March 10, 2011, 02:14:15 am »
It still is boolean
remember the calc only operates in 1's and 0's no matter what, this just acts like a whole bunch of boolean OR's for each bit in the number
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leafy
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Re: Graviter - Axe
«
Reply #263 on:
March 10, 2011, 02:16:30 am »
Alright, so now I just need to find the exact value that will jump exactly one block without hitting the ceiling.
Sounds like fun.
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Builderboy
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Re: Graviter - Axe
«
Reply #264 on:
March 10, 2011, 02:17:42 am »
I'm guessing 255 will do it, since that will put you at the lowest possible spot without moving into the next pixel, but i cant be sure o.O
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leafy
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Re: Graviter - Axe
«
Reply #265 on:
March 10, 2011, 02:23:20 am »
No my jumping works by setting a y-velocity to -value, so it adds up rather than just moving upwards by 255. So it's something like
x(x+1)/2>=255
or something, then use x.
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Builderboy
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Re: Graviter - Axe
«
Reply #266 on:
March 10, 2011, 02:25:21 am »
No i mean lowest possible spot on the screen (i've been talking about Y position) since higher numbers move down the screen
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leafy
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Re: Graviter - Axe
«
Reply #267 on:
March 10, 2011, 02:26:04 am »
Oh yeah 255 works for that. I got it working, so now I"m trying to figure out jumping.
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Builderboy
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Re: Graviter - Axe
«
Reply #268 on:
March 10, 2011, 02:26:54 am »
yay ^^
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leafy
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Seizon senryakuuuu!
Re: Graviter - Axe
«
Reply #269 on:
March 10, 2011, 02:36:04 am »
Wow I just realized I spammed this thread a whole lot >.< I'll stop now and go sleep or something.
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Graviter