Rust error E0382: borrow of moved value

The compiler team's explanation

A variable was used after its contents have been moved elsewhere.

Erroneous code example:

struct MyStruct { s: u32 }

fn main() {
    let mut x = MyStruct{ s: 5u32 };
    let y = x;
    x.s = 6;
    println!("{}", x.s);
}

Since MyStruct is a type that is not marked Copy, the data gets moved out of x when we set y. This is fundamental to Rust's ownership system: outside of workarounds like Rc, a value cannot be owned by more than one variable.

Sometimes we don't need to move the value. Using a reference, we can let another function borrow the value without changing its ownership. In the example below, we don't actually have to move our string to calculate_length, we can give it a reference to it with & instead.

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()
}

A mutable reference can be created with &mut.

Sometimes we don't want a reference, but a duplicate. All types marked Clone can be duplicated by calling .clone(). Subsequent changes to a clone do not affect the original variable.

Most types in the standard library are marked Clone. The example below demonstrates using clone() on a string. s1 is first set to "many", and then copied to s2. Then the first character of s1 is removed, without affecting s2. "any many" is printed to the console.

fn main() {
    let mut s1 = String::from("many");
    let s2 = s1.clone();
    s1.remove(0);
    println!("{} {}", s1, s2);
}

If we control the definition of a type, we can implement Clone on it ourselves with #[derive(Clone)].

Some types have no ownership semantics at all and are trivial to duplicate. An example is i32 and the other number types. We don't have to call .clone() to clone them, because they are marked Copy in addition to Clone. Implicit cloning is more convenient in this case. We can mark our own types Copy if all their members also are marked Copy.

In the example below, we implement a Point type. Because it only stores two integers, we opt-out of ownership semantics with Copy. Then we can let p2 = p1 without p1 being moved.

#[derive(Copy, Clone)]
struct Point { x: i32, y: i32 }

fn main() {
    let mut p1 = Point{ x: -1, y: 2 };
    let p2 = p1;
    p1.x = 1;
    println!("p1: {}, {}", p1.x, p1.y);
    println!("p2: {}, {}", p2.x, p2.y);
}

Alternatively, if we don't control the struct's definition, or mutable shared ownership is truly required, we can use Rc and RefCell:

use std::cell::RefCell;
use std::rc::Rc;

struct MyStruct { s: u32 }

fn main() {
    let mut x = Rc::new(RefCell::new(MyStruct{ s: 5u32 }));
    let y = x.clone();
    x.borrow_mut().s = 6;
    println!("{}", x.borrow().s);
}

With this approach, x and y share ownership of the data via the Rc (reference count type). RefCell essentially performs runtime borrow checking: ensuring that at most one writer or multiple readers can access the data at any one time.

If you wish to learn more about ownership in Rust, start with the Understanding Ownership chapter in the Book.

From rustc --explain E0382 (1.99.0), part of the Rust project, MIT or Apache-2.0. Official page.