I have a 2-dimensional vector:
let vec2d = vec![
vec![1, 1, 1],
vec![1, 1, 1],
];
I can yield the total of elements stored this way:
let mut n_vec_element: i32 = 0;
for i in vec2d.iter() {
n_vec_element += i.len() as i32;
}
println!("2D vector elements :{}", n_vec_element); // prints 6
When I increase the dimensions, the loop gets longer:
let mut n_vec_element: i32 = 0;
let vec3d = vec![
vec![
vec![1, 3, 5 as i32],
vec![2, 4, 6 as i32],
vec![3, 5, 7 as i32],
],
vec![
vec![1, 3, 5 as i32],
vec![2, 4, 6 as i32],
vec![3, 5, 7 as i32],
]
];
for i in vec3d.iter() {
// I must add another iter everytime I increment the dimension by 1.
// Else, it returns the number of stored vector instead of the vector
// elements.
for j in i.iter() {
n_vec_size += j.len() as i32;
}
};
println!("3D vector elements :{}", n_vec_element); // prints 18
There must be a more concise way to do this, but I haven't figured it out yet. Initially, I tried using vector's len() function but as I said above, it returns number of vectors stored instead of its elements.