Mesh m = Mesh.CreateFromSphere(new Sphere(Point3d.Origin, 6), 10, 10);
var list = new System.Collections.Generic.List(); list.Add(lc);
list.Add(m);
var formatter = new System.Runtime.Serialization.Formatters.Binary.BinaryFormatter(); using (var fs = new System.IO.FileStream(filename, System.IO.FileMode.Create))
{
formatter.Serialize(fs, list);
fs.Close();
}
using (var fs = new System.IO.FileStream(filename, System.IO.FileMode.Open))
{
object o = formatter.Deserialize(fs);
var retrieved_list = o as System.Collections.Generic.List;
fs.Close();
}
…
ta.
Take the following example.
I have two curves referenced with a Crv Param this creates a single path with 2 items {0}(N=2).
If I divide these Curves by 10 segments I end up with 2 Paths of 11 points each {0;0}(N=11) and {0;1}(N=11)
If I want each point to be handled separately I can Graft a branch to hold each point. So I get 22 Paths looking like {0;0;0}(N=1), {0;0;1}(N=1).....{0;1;10}(N=1).
Each path is therefore the address of a particular item or items of interest.
The Param Viewer can be found Params>Special>Param Viewer and can be viewed either as a list of Paths or graphically by a double click.
Also note that the Fancy Wires display the structure of a stream of data. Single wire = Single Data Item, Double Wire = Multiple Data on single Path and Double Dashed Wire = Multiple/Single Data on Multiple Paths…
king into the lists and and icant figure out the structure, i'ts differnet then the oreginal structure (10 items VS 11 items...).
eventually i want to create a list of curves and the number of intersections it has.
is it possible?…
nique name like in your example {0;0},{0;1},{0;2},{0;3} where each path contains 11 elements(curves,points,values,etc)
so if you want to restructure the structure first right click on path mapper and take create null mapping which reads your current path structure into the component
after that you can additional write on the left side to the existing path (i) as variable for the elements and then on the right side type in the structure you want to achieve
maybe {A;i}(B) so after that you have 11 paths where each have 3 elements
hope that helps…