[Rust] Collect()

String collect: automaticlly calling concat on string

    let foo: String = vec!["this", "is", "a", "test"]
        .into_iter()
        .collect();
    println!("{:?}", foo); // thisisatest

 

HashSet:

let foo: HashSet<isize> = vec![1, 2, 3]
    .into_iter()
    .collect();

 

HashMap:

use std::collections::HashMap;

fn main() {
    let foo: HashMap<&str, usize> = vec!["this", "is", "a", "test"]
        .into_iter()
        .enumerate()
        .map(|(idx, item)| (item, idx))
        .collect();
    println!("{:?}", foo); // {"this": 0, "is": 1, "a": 2, "test": 3}
}

 

Number:

let value: usize = vev![1,2,3]
    .iter()
    .sum(); // 6

 

let how_many_items: usize = vec![1,2,3]
    .iter()
    .skip(2)
    .count(); // 1

 

vec![1,2,5,9,4]
    .iter()
    .skip(2)
    .take_while(|&x| x > 4)
    .for_each(|&x| println!("{}", x)) // 5, 9

 

let what_about_this: usize = vec![1, 2, 3]
    .iter()
    .filter(|x| *x % 2 == 0)
    .count() // 1

In Rust, the .iter() method on a Vec<T> creates an iterator over immutable references to the elements of the vector. Therefore, in the context of your filter function, x is of type &i32, not i32.

The * operator is used to dereference x to get the value it points to. Without it, you'd be trying to perform the modulus operation on a reference (&i32), not the integer value (i32) itself. The Rust compiler will not automatically dereference the value for you in this context, so you need to use the * operator to explicitly dereference it.

 

let map = HashMap::from([
   ("foo", 1),
   ("bar", 2),
   ("baz", 3),
]);

map
    .iter()
    .for_each(|(k, v)| println!("{}: {}", k, v));




let set = HashSet::from([
    "foo",
    "bar",
    "baz",
]);

set
    .iter()
    .for_each(|v| println!("{}", v));