I'm trying to transform some points in a point wrangle and
vector vec = set(1,1,1);
@P += vec;
@P -= vec;
returns points to their starting position as expected, but
vector vec = set(1,1,1);
@P = point(0, “P”, @ptnum)+vec;
@P = point(0, “P”, @ptnum)-vec;
gives different results, as if the first transform didn't do anything.
why are the results different? am I using point() wrong? sorry if this is too common a question.
@P vs Point() problem
5725 6 1-
- xavier_h
- Member
- 4 posts
- Joined: Feb. 2017
- Offline
-
- Sean Rowe
- Member
- 39 posts
- Joined: July 2013
- Offline
-
- xavier_h
- Member
- 4 posts
- Joined: Feb. 2017
- Offline
If that's the case, then why does
vector vec = set(1,1,1);
@P = @P+vec;
@P = @P-vec;
work the same way as the original
vector vec = set(1,1,1);
@P += vec;
@P -= vec;
yet acts different than
vector vec = set(1,1,1);
@P = point(0, “P”, @ptnum)+vec;
@P = point(0, “P”, @ptnum)-vec;
don't @P and point(0, “P”, @ptnum) do the same thing?
vector vec = set(1,1,1);
@P = @P+vec;
@P = @P-vec;
work the same way as the original
vector vec = set(1,1,1);
@P += vec;
@P -= vec;
yet acts different than
vector vec = set(1,1,1);
@P = point(0, “P”, @ptnum)+vec;
@P = point(0, “P”, @ptnum)-vec;
don't @P and point(0, “P”, @ptnum) do the same thing?
-
- Sean Rowe
- Member
- 39 posts
- Joined: July 2013
- Offline
They do mean the same thing, if you bound both ways to two new attributes they would match.
I think the reason that the top expressions work is because of the way code is read in general coding practices.
@P = @P+vec is recognised as @P += vec. The point expression being a vex function probably isn't recognised the same way.
I think the reason that the top expressions work is because of the way code is read in general coding practices.
@P = @P+vec is recognised as @P += vec. The point expression being a vex function probably isn't recognised the same way.
-
- tamte
- Member
- 8548 posts
- Joined: July 2007
- Online
you have to imagine it step by step:
the same goes for @P += and -= example as that's just the shorter syntax to the same assignment
but in here:
also worth noting that point(0, “P”, @ptnum) is pretty ambiguous andin this case Houdini makes a good guess you want to use vector representation, but if using within a longer expression and not just on it's own as a part of assignment it's safer to use vector(point(0, “P”, @ptnum))
vector vec = set(1,1,1); // imagine @P is {0,0,0} @P = @P+vec; // now you are altering @P to be {0,0,0} + {1,1,1} = {1,1,1} // so from now on @P is no longer {0,0,0} but {1,1,1} @P = @P-vec; // now you are alreting @P to be {1,1,1} - {1,1,1} so back to {0,0,0}
the same goes for @P += and -= example as that's just the shorter syntax to the same assignment
but in here:
vector vec = set(1,1,1); // imagine @P is {0,0,0} @P = point(0, “P”, @ptnum)+vec; // now you are altering @P to be {0,0,0} + {1,1,1} = {1,1,1} // so from now on @P is no longer {0,0,0} but {1,1,1} // however point(0, “P”, @ptnum) function takes P attrib from the input 0 so it will always return {0,0,0} // even after this point as it has no connection to @P @P = point(0, “P”, @ptnum)-vec; // therefore here you are setting @P to be again {0,0,0} - {1,1,1} so {-1,-1,-1}
also worth noting that point(0, “P”, @ptnum) is pretty ambiguous andin this case Houdini makes a good guess you want to use vector representation, but if using within a longer expression and not just on it's own as a part of assignment it's safer to use vector(point(0, “P”, @ptnum))
Tomas Slancik
FX Supervisor
Method Studios, NY
FX Supervisor
Method Studios, NY
-
- vusta
- Member
- 555 posts
- Joined: Feb. 2017
- Offline
I'm not technical so pls don't hang me for this…but I'll try:
if you use
@P += vec;
@P -= vec;
then you are using the updated pos in the wrangle.
if you use
@P = point(0, “P”, @ptnum)+vec;
@P = point(0, “P”, @ptnum)-vec;
then the point() refers to what you STARTED with before you got into this wrangle.
So use this test:
vector vec = set(1,1,1);
@P += vec;
printf(“y:%f\n”,vector(point(0, “P”, @ptnum)).y);
printf(“Py:%f\n”, @P.y);
and you'll see clearly @P inside the wrangle is updated…but point() is not using the @P inside the wrangle. It's using the orig input before any manipulation.
if you use
@P += vec;
@P -= vec;
then you are using the updated pos in the wrangle.
if you use
@P = point(0, “P”, @ptnum)+vec;
@P = point(0, “P”, @ptnum)-vec;
then the point() refers to what you STARTED with before you got into this wrangle.
So use this test:
vector vec = set(1,1,1);
@P += vec;
printf(“y:%f\n”,vector(point(0, “P”, @ptnum)).y);
printf(“Py:%f\n”, @P.y);
and you'll see clearly @P inside the wrangle is updated…but point() is not using the @P inside the wrangle. It's using the orig input before any manipulation.
Edited by vusta - March 8, 2018 20:18:47
-
- xavier_h
- Member
- 4 posts
- Joined: Feb. 2017
- Offline
-
- Quick Links