beginner VEX question / set @Cd with custom value does not work

[jon3de]

Hi,

For educational purposes I try to replace things I can do with point SOPs with small VEX codes in attributeWrangle. I do that to get a little bit warm with the language also I read that it computes faster than a regular point SOP. ( Not that it would be necessary for my small learning scenes but....anyway)

 

I created  a attribute called "displaceValue" and wanted to use this to color the CD attribute. But there has to be a mistake in the code and I can´t find it. 

The attached picture explains it better. I would be thankful for some help.

I also tried something like " @Cd = {(@ptnum/@numpt),0,0};" which also does not work.

 

 

kind regards 

Jonathan

 

EDIT:

 

Ah....it works with:

 

@displaceValue = float(@ptnum)/float(@numpt);

 

@Cd = set(@displaceValue,0,0);

 

So i think the curly brackets caused the error...

Are the curly brackets just for fast "values" and no attributes or variables?

If someone wants to explain the error in more detail to me I would appreciate that.

Edited October 23, 2015 by jon3de

[jkunz07]

You can't have variables inside of brackets like that.

 

You would have to do:

 

@Cd = set(@displaceValue, 0, 0);

[mestela]

Another way which can sometimes be easier to read is dot syntax:

 

@Cd = 0;

@Cd.r = @displaceValue;

 

@Cd is one of the few special values that automatically converts itself to a vector, so @Cd=0; gets converted to @Cd={0,0,0};

 

You can also call the suffix as r/g/b, or x/y/z, or an array index, all are useful depending on circumstance. Ie,

 

@Cd.x = 1;

@Cd.r  = 1;

@Cd[0] = 1;

 

are all the same, and will make the points red component to 1.

Edited October 23, 2015 by mestela

[jon3de]

Thanks a lot for the additional info. 

By the way...I always get a zero value  if I write the @ptnum and @numpt without the "float" is that because that are integers by default?

And is the way I converted them the right way? Or is there a better way to handle this? Thanks again for all the help.

Jon

[f1480187]

You are right.

int a=3, b=2;
i@i = a / b;
f@f = (float) a / b;

No need to cast both sides of expression to float, by the way. The right side will be converted implicitly. There will be an Implicit Bingo in single wrangle.

[jon3de]

Ah got it...

So with your help I have now:

 

@displaceValue = (float) @ptnum/@numpt;

 

@Cd = 0;

@Cd.r = fit01(@displaceValue,1,0);

 

@P.y += @Cd.r*3.5;

 

it works just fine!