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Messages - nemo
Pages: 1 ... 70 71 [72] 73 74 ... 82
1066
« on: June 27, 2010, 01:07:11 am »
no the width does not have to be specified as a multiple of eight, but when converting the sprite to hex you have to pretend that the width is a multiple of 8. so if you had a 9x9 black square:
111111111
111111111
111111111
111111111
111111111
111111111
111111111
111111111
111111111
you would specify the dimensions as 0909 in the first two bytes. however, the sprite you convert to hex would look like so:
1111111110000000
1111111110000000
1111111110000000
1111111110000000
1111111110000000
1111111110000000
1111111110000000
1111111110000000
1111111110000000
because you have to pad to the nearest byte, or multiple of eight.
the hex for a 9x9 square would thus be:
[0909FF80FF80FF80FF80FF80FF80FF80FF80FF80]
1067
« on: June 26, 2010, 08:08:14 pm »
i'd vote for string output if that was an option.
edit: would it be feasible to have a shortcut to incrementing a byte in RAM?
so rather than doing
{O+1*Y+X+L1}+1->{O+1*Y+X+L1}
you could do something like
{O+1*Y+X+L1}++
just like in C.
1068
« on: June 26, 2010, 07:21:30 pm »
i hope it is too.. that would come REALLY handy in my contest entry (: i'm waiting to place my vote until i know though.
1069
« on: June 26, 2010, 06:51:17 pm »
will drawing a 8x8 black square be faster or slower than using an 8x8 sprite?
1070
« on: June 26, 2010, 06:30:07 pm »
000110010
000111000
011000110
100101001
101010101
100000001
101000101
100111001
011000110
000111000
is [090919001C0063009480AA808080A2809C8063001C00]
i just tried it on calc and it works.
1071
« on: June 26, 2010, 05:59:10 pm »
i think that's wrong, but i'm not completely sure. it's 9x9 so wouldn't 0909 be the first two bytes in the hex?
1072
« on: June 26, 2010, 05:05:30 pm »
i think negative values are easy to understand, just tough to implement. also, getting sin cos and tan are difficult. i still don't understand bitmap(). but i don't use it either.
1073
« on: June 26, 2010, 12:57:27 am »
ohhh i got you. yeah i think it's because each paranthesis/brackets/braces counts as a certain amount of memory in the stack. so {L1} and {L1+12} take up the same amount of memory in the stack and therefore are equal in size.
1074
« on: June 26, 2010, 12:52:35 am »
wait.. i'm lost by what you guys mean by "size". all expressions within braces ({}) are one byte... and {}r is two bytes..
1075
« on: June 26, 2010, 12:46:07 am »
{} gets the byte the pointer points to. 0->{L1}:Disp {L1}>Dec. 0 would be displayed.
1. yes.
2. depends on your order of operations.
3. i think so, yes.
4. i think so, yes
1076
« on: June 24, 2010, 11:22:09 pm »
no, this is 6Mhz.
1077
« on: June 24, 2010, 05:22:17 pm »
yes. i know from experience.
1078
« on: June 24, 2010, 05:05:04 pm »
i'm not sure DJ, i tried putting a Pause 50 at the beginning of each If statement to no avail but that didn't work either. also, if you notice it's not that the keypresses aren't being detected, it's just either a) the wrong value is being stored into S or b) the wrong keypress is being detected. because that code always exits if you press one of the keys 4-8, meaning it IS registering the fact that a key has been pressed.
1079
« on: June 24, 2010, 04:44:22 pm »
that's my favorite level. i remember doing it in "advanced" mode, where all the bots are in cages so you can't kill them... but if you're trying to figure out how to, DJ, if you looked above the bots there was usually a place you could shoot a portal. then you'd make a different portal elsewhere and drop a cube through it, onto the bot.
builderboy: if a player picks up a cube, then walks to the edge so that the cube is over the edge, will the cube fall down? especially if the player is positioned like so, with just 1 pixel of his leg on the edge of a cliff:
PLAYER| CUBE
|_ ____
_|_ | |
_____________| | |___|
Cliff |
|
1080
« on: June 24, 2010, 04:36:19 pm »
yeah, you could push boxes around in the original portal, but it's more logical and convenient to pick them up. ahh.. the good days of the companion cube being used as a shield (:
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