What is the difference between Copy and Clone?

Question

This issue seems to imply it's just an implementation detail (memcpy vs ???), but I can't find any explicit description of the differences.

Answer 1

Clone is designed for arbitrary duplications: a Clone implementation for a type T can do arbitrarily complicated operations required to create a new T. It is a normal trait (other than being in the prelude), and so requires being used like a normal trait, with method calls, etc.

The Copy trait represents values that can be safely duplicated via memcpy: things like reassignments and passing an argument by-value to a function are always memcpys, and so for Copy types, the compiler understands that it doesn't need to consider those a move.

Answer 2

The main difference is that cloning is explicit. Implicit notation means move for a non-Copy type.

// u8 implements Copy
let x: u8 = 123;
let y = x;
// x can still be used
println!("x={}, y={}", x, y);

// Vec<u8> implements Clone, but not Copy
let v: Vec<u8> = vec![1, 2, 3];
let w = v.clone();
//let w = v // This would *move* the value, rendering v unusable.

By the way, every Copy type is also required to be Clone. However, they are not required to do the same thing! For your own types, .clone() can be an arbitrary method of your choice, whereas implicit copying will always trigger a memcpy, not the clone(&self) implementation.

Answer 3

As already covered by other answers:

• Copy is implicit, inexpensive, and cannot be re-implemented (memcpy).
• Clone is explicit, may be expensive, and may be re-implement arbitrarily.

What is sometimes missing in the discussion of Copy vs Clone is that it also affects how the compiler uses moves vs automatic copies. For instance:

#[derive(Debug, Clone, Copy)]
pub struct PointCloneAndCopy {
    pub x: f64,
}

#[derive(Debug, Clone)]
pub struct PointCloneOnly {
    pub x: f64,
}

fn test_copy_and_clone() {
    let p1 = PointCloneAndCopy { x: 0. };
    let p2 = p1; // because type has `Copy`, it gets copied automatically.
    println!("{:?} {:?}", p1, p2);
}

fn test_clone_only() {
    let p1 = PointCloneOnly { x: 0. };
    let p2 = p1; // because type has no `Copy`, this is a move instead.
    println!("{:?} {:?}", p1, p2);
}

(Rust Playground)

The first example (PointCloneAndCopy) works fine here because of the implicit copy, but the second example (PointCloneOnly) would error with a use after move:

error[E0382]: borrow of moved value: `p1`
  --> src/lib.rs:20:27
   |
18 |     let p1 = PointCloneOnly { x: 0. };
   |         -- move occurs because `p1` has type `PointCloneOnly`, which does not implement the `Copy` trait
19 |     let p2 = p1;
   |              -- value moved here
20 |     println!("{:?} {:?}", p1, p2);
   |                           ^^ value borrowed here after move

To avoid the implicit move, we could explicitly call let p2 = p1.clone();.

This may raise the question of how to force a move of a type which implements the Copy trait?.

Short answer: You can't / doesn't make sense.

Answer 4

As written here.

Copies happen implicitly, for example as part of an assignment y = x. The behavior of Copy is not overloadable; it is always a simple bit-wise copy.

Cloning is an explicit action, x.clone(). The implementation of Clone can provide any type-specific behavior necessary to duplicate values safely. For example, the implementation of Clone for String needs to copy the pointed-to string buffer in the heap. A simple bitwise copy of String values would merely copy the pointer, leading to a double free down the line. For this reason, String is Clone but not Copy.

Clone is a supertrait of Copy, so everything which is Copy must also implement Clone. If a type is Copy then its Clone implementation only needs to return *self

Answer 5

IMHO, the intention is all your need

• both can be used to avoid borrow and lifetime annotation or smart pointer
• copy means copy semantics here, or means the value is going to be treated like in Java or JS language, this implies the value is somehow primitive or cheap to copy
• clone let you decide how deep the duplication is enough and you have to explicitly write .clone, which means it is already handled

Answer 6

I found this explanation very helpful:

In Rust, some simple types are “implicitly copyable” and when you assign them or pass them as arguments, the receiver will get a copy, leaving the original value in place. For other types, copies must be made explicitly, by implementing the Clone trait and calling the clone() method.

The Clone trait defines the ability to explicitly create a deep copy of an object T. When we call Clone for type T, it does all the arbitrarily complicated operations required to create a new T.

The Copy trait in rust defines the ability to implicitly copy an object. The behavior Copy is not overloadable. It is always a simple bitwise copy. This is available for the types that have a fixed size and are stored entirely on the stack.

Reference: https://intmain.co/difference-between-copy-and-clone-trait-in-rust