JFASI

11th November 2005 17:02 UTC

where am i going wrong???
Ok, so I got a problem. Something i'm making involves creating random quadrilaterals.

Here's my scope code:

Init

n=5

Frame

nr=0

Beat

cx1=rand(100)/100-.5;
cx2=rand(100)/100-.5;
cx3=rand(100)/100-.5;
cx4=rand(100)/100-.5;

cy1=rand(100)/100-.5;
cy2=rand(100)/100-.5;
cy3=rand(100)/100-.5;
cy4=rand(100)/100-.5;

Point

nr=nr+1;

x1=if(equals(nr,1),cx1,x1);
x1=if(equals(nr,2),cx2,x1);
x1=if(equals(nr,3),cx3,x1);
x1=if(equals(nr,4),cx4,x1);

y1=if(equals(nr,1),cy1,y1);
y1=if(equals(nr,2),cy2,y1);
y1=if(equals(nr,3),cy3,y1);
y1=if(equals(nr,4),cy4,y1);

x=x1;y=y1


no matter what I do, it flops. This isn't a one shot problem. I'm probably doing something conceptually wrong, but I can't find it.

Tuggummi

11th November 2005 18:48 UTC

*ahem* "equals"?

eheiney

11th November 2005 18:50 UTC

I'll give you one little hint. equals() is not a function. ;)

JFASI

11th November 2005 19:17 UTC

Ooooooops.


:confused: The one thing I didn't check, either :confused:


Thanks





Now how can I extend this to more than 4 points...

Is there anything in the syntax that would mean:

cx(nr)

?

Hmmmm. I could take the code out of the Beat, and use the b variable...

UIUC85

11th November 2005 23:00 UTC

Theres no way to make new arrays like what you're describing. However if you look at the help you do have megabuf() and loop() at your command. So you could try something like...

init:
n=8;

frame:
nr=0;

beat:
loop(2*n,exec2(assign(nr,nr+1),assign(megabuf(nr),rand(100)/100-.5)));
assign(megabuf(n),megabuf(1));
assign(megabuf(n+1),megabuf(2*n));

point:
nr=nr+1;
x=megabuf(nr);
y=megabuf(nr+n);

Now you can change n to whatever you want and get a random shape with n-1 sides!

JFASI

14th November 2005 03:01 UTC

Ah... I think I get loop better. Does it force the section to be repeated for whatever amount of time you make it? That would make things easier for me...

UIUC85

14th November 2005 04:59 UTC

yeah it'll execute the statement it is paired with that many times or in this case 2*n. Since there's n x values and n y values then you need 2*n values and to close off the shape you have the last and first values of x's and y's be the same which is why the statements below it are there.