pub type RowSVector<T, const D: usize> = Matrix<T, U1, Const<D>, ArrayStorage<T, 1, D>>;Expand description
A statically sized D-dimensional row vector.
Aliased Type§
struct RowSVector<T, const D: usize> {
pub data: ArrayStorage<T, 1, D>,
/* private fields */
}Fields§
§data: ArrayStorage<T, 1, D>The data storage that contains all the matrix components. Disappointed?
Well, if you came here to see how you can access the matrix components,
you may be in luck: you can access the individual components of all vectors with compile-time
dimensions <= 6 using field notation like this:
vec.x, vec.y, vec.z, vec.w, vec.a, vec.b. Reference and assignation work too:
let mut vec = Vector3::new(1.0, 2.0, 3.0);
vec.x = 10.0;
vec.y += 30.0;
assert_eq!(vec.x, 10.0);
assert_eq!(vec.y + 100.0, 132.0);Similarly, for matrices with compile-time dimensions <= 6, you can use field notation
like this: mat.m11, mat.m42, etc. The first digit identifies the row to address
and the second digit identifies the column to address. So mat.m13 identifies the component
at the first row and third column (note that the count of rows and columns start at 1 instead
of 0 here. This is so we match the mathematical notation).
For all matrices and vectors, independently from their size, individual components can
be accessed and modified using indexing: vec[20], mat[(20, 19)]. Here the indexing
starts at 0 as you would expect.