What is the difference between ref and &??

Today I encountered the use of ref while learning the code to use enumerations to implement linked lists, so I would like to ask, what is the difference between ref and &? When should we use ref instead of &?

use crate::List::*;

enum List {
    // Cons: Tuple struct that wraps an element and a pointer to the next node
    Cons(u32, Box<List>),
    // Nil: A node that signifies the end of the linked list

// Methods can be attached to an enum
impl List {
    // Create an empty list
    fn new() -> List {
        // `Nil` has type `List`

    // Consume a list, and return the same list with a new element at its front
    fn prepend(self, elem: u32) -> List {
        // `Cons` also has type List
        Cons(elem, Box::new(self))

    // Return the length of the list
    fn len(&self) -> u32 {
        // `self` has to be matched, because the behavior of this method
        // depends on the variant of `self`
        // `self` has type `&List`, and `*self` has type `List`, matching on a
        // concrete type `T` is preferred over a match on a reference `&T`
        // after Rust 2018 you can use self here and tail (with no ref) below as well,
        // rust will infer &s and ref tail. 
        // See https://doc.rust-lang.org/edition-guide/rust-2018/ownership-and-lifetimes/default-match-bindings.html
        match *self {
            // Can't take ownership of the tail, because `self` is borrowed;
            // instead take a reference to the tail
            Cons(_, ref tail) => 1 + tail.len(),
            // Base Case: An empty list has zero length
            Nil => 0

    // Return representation of the list as a (heap allocated) string
    fn stringify(&self) -> String {
        match *self {
            Cons(head, ref tail) => {
                // `format!` is similar to `print!`, but returns a heap
                // allocated string instead of printing to the console
                format!("{}, {}", head, tail.stringify())
            Nil => {

fn main() {
    // Create an empty linked list
    let mut list = List::new();

    // Prepend some elements
    list = list.prepend(1);
    list = list.prepend(2);
    list = list.prepend(3);

    // Show the final state of the list
    println!("linked list has length: {}", list.len());
    println!("{}", list.stringify());

This is a common point of confusion. In simple words:

  • In expressions, &/&mut creates a reference and * dereferences it (dereferencing is also done automatically in several cases).
  • In patterns, &/&mut "matches" (i.e. dereferences) a reference, and ref/ref mut creates it.

An article which helped me understand this question: Patterns Are Not Expressions


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