Index

References and Borrowing

References and Borrowing

One case use reference without moving the string to calculate_length

fn main() {
    let s1 = String::from("hello");

    let len = calculate_length(&s1);

    println!("The length of '{}' is {}.", s1, len);
}

fn calculate_length(s: &String) -> usize {
    s.len()
}

Note that we pass &s1 into calculate_length and, in its definition, we take &String rather than String

These ampersands are references, and they allow you to refer to some value without taking ownership of it.

images/871-1.png

Note: The opposite of referencing by using & is dereferencing, which is accomplished with the dereference operator, *

let s1 = String::from("hello");

let len = calculate_length(&s1);

The &s1 syntax lets us create a reference that refers to the value of s1 but does not own it. Because it does not own it, the value it points to will not be dropped when the reference goes out of scope.

fn calculate_length(s: &String) -> usize { // s is a reference to a String
s.len()
} // Here, s goes out of scope. But because it does not have ownership of what
// it refers to, nothing happens.

We call having references as function parameters borrowing. As in real life, if a person owns something, you can borrow it from them. When you’re done, you have to give it back.

But mutable references have one big restriction: you can have only one mutable reference to a particular piece of data in a particular scope. This code will fail:
Error code:

    let mut s = String::from("hello");

    let r1 = &mut s;
    let r2 = &mut s;

    println!("{}, {}", r1, r2);

As always, we can use curly brackets to create a new scope, allowing for multiple mutable references, just not simultaneous ones:

    let mut s = String::from("hello");

    {
        let r1 = &mut s;
    } // r1 goes out of scope here, so we can make a new reference with no problems.

    let r2 = &mut s;

A similar rule exists for combining mutable and immutable references. This code results in an error:
Error code:

    let mut s = String::from("hello");

    let r1 = &s; // no problem
    let r2 = &s; // no problem
    let r3 = &mut s; // BIG PROBLEM

    println!("{}, {}, and {}", r1, r2, r3);

Note that a reference’s scope starts from where it is introduced and continues through the last time that reference is used

fn main() {
    let mut s = String::from("hello");

    let r1 = &s; // no problem
    let r2 = &s; // no problem
    println!("{} and {}", r1, r2);
    // r1 and r2 are no longer used after this point

    let r3 = &mut s; // no problem
    println!("{}", r3);
}

Dangling References

In languages with pointers, it’s easy to erroneously create a dangling pointer, a pointer that references a location in memory that may have been given to someone else, by freeing some memory while preserving a pointer to that memory.

In Rust, by contrast, the compiler guarantees that references will never be dangling references: if you have a reference to some data, the compiler will ensure that the data will not go out of scope before the reference to the data does.

Error code:

fn main() {
    let reference_to_nothing = dangle();
}

fn dangle() -> &String {
    let s = String::from("hello");

    &s
}

Error code:

fn dangle() -> &String { // dangle returns a reference to a String

    let s = String::from("hello"); // s is a new String

    &s // we return a reference to the String, s
} // Here, s goes out of scope, and is dropped. Its memory goes away.
  // Danger!

Because s is created inside dangle, when the code of dangle is finished, s will be deallocated. But we tried to return a reference to it. That means this reference would be pointing to an invalid String. That’s no good! Rust won’t let us do this.

The solution here is to return the String directly:

fn main() {
    let string = no_dangle();
}

fn no_dangle() -> String {
    let s = String::from("hello");

    s
}