Rust The dot operator


Example

The . operator in Rust comes with a lot of magic! When you use ., the compiler will insert as many *s (dereferencing operations) necessary to find the method down the deref "tree". As this happens at compile time, there is no runtime cost of finding the method.

let mut name: String = "hello world".to_string();
// no deref happens here because push is defined in String itself
name.push('!');

let name_ref: &String = &name;
// Auto deref happens here to get to the String. See below
let name_len = name_ref.len();
// You can think of this as syntactic sugar for the following line:
let name_len2 = (*name_ref).len();

// Because of how the deref rules work,
// you can have an arbitrary number of references. 
// The . operator is clever enough to know what to do.
let name_len3 = (&&&&&&&&&&&&name).len();
assert_eq!(name_len3, name_len);

Auto dereferencing also works for any type implementing std::ops::Deref trait.

let vec = vec![1, 2, 3];
let iterator = vec.iter();

Here, iter is not a method of Vec<T>, but a method of [T]. It works because Vec<T> implements Deref with Target=[T] which lets Vec<T> turn into [T] when dereferenced by the * operator (which the compiler may insert during a .).