View Full Version : polycount?
ginodauri 09-19-2008, 04:35 PM How to get polycount?
I dont understand.When i want to declare variable is must be array , like:
int $polygonCount[] =` polyEvaluate -f `
size of this array is 1.
I need just int variable of polycount.
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greatPumpkin
09-19-2008, 05:46 PM
the size of the array is not the number of faces,
int $polygonCount[] =` polyEvaluate -f `;
so your array size is one.. but your actual face count is:
print $polygonCount[0];
maya does this because you can pass multiple flags to polyEvalute, so if you pass -f -v, your first array item will be number of faces, and your second item will be number of verts.
ginodauri
09-20-2008, 03:24 AM
Ok that is clear.
I have some problem with this code,like infinite loop.
Someone please tell me what provlem is.
It is for random extrude poligons of poligonal object.
int $polygonCount[] =` polyEvaluate -f `;
int $currentCount = $polygonCount[0];
int $randomFace;
float $extrudeFace1;
int $counter;
string $selection [] = `ls -sl';
for ($each in $selection)
{
for ($counter=0; $counter<$currentCount; $counter++)
{
int $randomFace = `rand $currentCount`;
float $extrudeFace1 = $randomFace/10.0;
polyExtrudeFacet -kft false -ltz $extrudeFace1 $each.f[$randomFace];
}
}
Robert Bateman
09-22-2008, 10:44 AM
I have some problem with this code,like infinite loop.
When you say "like infinite loop", do you mean "doesn't run at all"?
polyExtrudeFacet -kft false -ltz $extrudeFace1 ($each+".f["+$randomFace+"]");
If you do mean "infinite loop", it makes me think you're putting this in an expression? which is bad news! don't create new nodes inside an expression node - since it changes the DG during evaluation, which will only make things go very wrong indeed....
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09-22-2008, 10:44 AM
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