如何克隆 Vec <Box<dyn Trait> &gt;?

Question

I want to implement a function that takes an immutable &Vec reference, makes a copy, sorts the values and prints them.

This is the main code.

trait Foo {
    fn value(&self) -> i32;
}

struct Bar {
    x: i32,
}

impl Foo for Bar {
    fn value(&self) -> i32 {
        self.x
    }
}

fn main() {
    let mut a: Vec<Box<dyn Foo>> = Vec::new();
    a.push(Box::new(Bar { x: 3 }));
    a.push(Box::new(Bar { x: 5 }));
    a.push(Box::new(Bar { x: 4 }));

    let b = &a;
    sort_and_print(&b);
}

The only way I was able to make it work was this

fn sort_and_print(v: &Vec<Box<dyn Foo>>) {
    let mut v_copy = Vec::new();
    for val in v {
        v_copy.push(val);
    }
    v_copy.sort_by_key(|o| o.value());
    for val in v_copy {
        println!("{}", val.value());
    }
}

However I want to understand what's happening here and also to make the code shorter.

Question 1

If I try to change let mut v_copy = Vec::new(); to let mut v_copy: Vec<Box<dyn Foo>> = Vec::new(); however that results in various errors that I don't know how to fix.

How do I make this version work and why is it different than the first version?

Attempt 2

Something closer to what I'm looking for is something like this. let mut v_copy = v.clone(); but this doesn't work. Can this version be fixed?

Answer 1

First, let's annotate the types:

fn sort_and_print(v: &Vec<Box<dyn Foo>>) {
    let mut v_copy: Vec<&Box<dyn Foo>> = Vec::new();
    for val /* : &Box<dyn Foo> */ in v {
        v_copy.push(val);
    }
    v_copy.sort_by_key(|o: &&Box<dyn Foo>| <dyn Foo>::value(&***o));
    for val /* : &Box<dyn Foo> */ in v_copy {
        println!("{}", <dyn Foo>::value(&**val));
    }
}

Iterating over &Vec<T> produces an iterator over &T (the same as the .iter() method.

Now we can see we can convert it into iterator, by either calling .into_iter() on v and then .collect() (which is what the for loop does), or replace into_iter() with iter() (which is more idiomatic since we're iterating over references:

fn sort_and_print(v: &Vec<Box<dyn Foo>>) {
    let mut v_copy: Vec<&Box<dyn Foo>> = v.iter().collect(); // You can omit the `&Box<dyn Foo>` and replace it with `_`, I put it here for clarity.
    v_copy.sort_by_key(|o| o.value());
    for val in v_copy {
        println!("{}", val.value());
    }
}

However, we still only have a copy of the reference ( &Box<dyn Foo> ). Why can't we just clone the vector?

If we try...

fn sort_and_print(v: &Vec<Box<dyn Foo>>) {
    let mut v_copy = v.clone();
    v_copy.sort_by_key(|o| o.value());
    for val in v_copy {
        println!("{}", val.value());
    }
}

...the compiler yell at us:

warning: variable does not need to be mutable
  --> src/main.rs:45:9
   |
45 |     let mut v_copy = v.clone();
   |         ----^^^^^^
   |         |
   |         help: remove this `mut`
   |
   = note: `#[warn(unused_mut)]` on by default

error[E0596]: cannot borrow `*v_copy` as mutable, as it is behind a `&` reference
  --> src/main.rs:46:5
   |
45 |     let mut v_copy = v.clone();
   |         ---------- help: consider changing this to be a mutable reference: `&mut Vec<Box<dyn Foo>>`
46 |     v_copy.sort_by_key(|o| o.value());
   |     ^^^^^^ `v_copy` is a `&` reference, so the data it refers to cannot be borrowed as mutable

WHAT???????????

Well, let's try to specify the type. It can make the compiler smarter.

fn sort_and_print(v: &Vec<Box<dyn Foo>>) {
    let mut v_copy: Vec<Box<dyn Foo>> = v.clone();
    v_copy.sort_by_key(|o| o.value());
    for val in v_copy {
        println!("{}", val.value());
    }
}

Nope.

error[E0308]: mismatched types
  --> src/main.rs:45:41
   |
45 |     let mut v_copy: Vec<Box<dyn Foo>> = v.clone();
   |                     -----------------   ^^^^^^^^^
   |                     |                   |
   |                     |                   expected struct `Vec`, found reference
   |                     |                   help: try using a conversion method: `v.to_vec()`
   |                     expected due to this
   |
   = note: expected struct `Vec<Box<dyn Foo>>`
           found reference `&Vec<Box<dyn Foo>>`

Well, let's use the compiler's suggestion:

fn sort_and_print(v: &Vec<Box<dyn Foo>>) {
    let mut v_copy: Vec<Box<dyn Foo>> = v.to_vec();
    v_copy.sort_by_key(|o| o.value());
    for val in v_copy {
        println!("{}", val.value());
    }
}

Grrr!!

error[E0277]: the trait bound `dyn Foo: Clone` is not satisfied
  --> src/main.rs:45:43
   |
45 |     let mut v_copy: Vec<Box<dyn Foo>> = v.to_vec();
   |                                           ^^^^^^ the trait `Clone` is not implemented for `dyn Foo`
   |
   = note: required because of the requirements on the impl of `Clone` for `Box<dyn Foo>`

At least we now have some clues.

What happened here?

Well, like the compiler said, dyn Foo does not implement the Clone trait. Which means that neither does Box<dyn Foo> , and so is Vec<Box<dyn Foo>> .

However, &Vec<Box<dyn Foo>> actually does impl Clone . This is because you can have as many shared references as you want - shared (non-mutable) references are Copy , and every Copy is also Clone . Try it:

fn main() {
    let i: i32 = 123;
    let r0: &i32 = &i;
    let r1: &i32 = <&i32 as Clone>::clone(&r0);
}

So, when we write v.clone() , the compiler asks "is there a method named clone() that takes self of type &Vec<Box<dyn Foo>> ( v )?" it first looks for such method on the Clone impl for Vec<Box<dyn Foo>> (because the Clone::clone() takes &self , so for Vec<Box<dyn Foo>> it takes &Vec<Box<dyn Foo>> ). Unfortunately, such impl doesn't exist, so it does the magic of autoref (part the process of trying to adjust a method receiver in Rust, you can read more here ), and asks the same question for &&Vec<Box<dyn Foo>> . Now we did find a match - <&Vec<Box<dyn Foo>> as Clone>::clone() ! So this is what the compiler calls.

What is the return type of the method? Well, &Vec<Box<dyn Foo>> . This will be the type of v_copy . Now we understand why when we specified another type, the compiler got crazy! We can also decrypt the error message when we didn't specify a type: we asked the compiler to call sort_by_key() on a &Vec<Box<dyn Foo>> , but this method requires a &mut Vec<Box<dyn Foo>> ( &mut [Box<dyn Foo>] , to be precise, but it doesn't matter because Vec<T> can coerce to [T] )!

We can also understand the warning about a redundant mut : we never change the reference , so no need to declare it as mutable!

When we called to_vec() , OTOH, the compiler didn't get confused: to_vec() requires the vector's item to implement Clone ( where T: Clone ), which doesn't happen for Box<dyn Foo> . BOOM.

Now the solution should be clear: we just need Box<dyn Foo> to impl Clone .

Just?...

The first thing we may think about is give Foo a supertrait Clone :

trait Foo: Clone {
    fn value(&self) -> i32;
}

#[derive(Clone)]
struct Bar { /* ... */ }

Not working:

error[E0038]: the trait `Foo` cannot be made into an object
  --> src/main.rs:33:31
   |
33 | fn sort_and_print(v: &Vec<Box<dyn Foo>>) {
   |                               ^^^^^^^ `Foo` cannot be made into an object
   |
note: for a trait to be "object safe" it needs to allow building a vtable to allow the call to be resolvable dynamically; for more information visit <https://doc.rust-lang.org/reference/items/traits.html#object-safety>
  --> src/main.rs:1:12
   |
1  | trait Foo: Clone {
   |       ---  ^^^^^ ...because it requires `Self: Sized`
   |       |
   |       this trait cannot be made into an object...

Hmm, looking like indeed Clone requires Sized . Why?

Well, because to clone something, we need to return itself. Can we return dyn Foo ? No, because it can be of any size.

So, let's try to impl Clone for Box<dyn Foo> by hand (we can do that even though Box is not defined in our crate because it is a fundamental type).

impl Clone for Box<dyn Foo> {
    fn clone(self: &Box<dyn Foo>) -> Box<dyn Foo> {
        // Now... What, actually?
    }
}

How can we even clone something that can be anything? Clearly we need to forward it to someone else. Who else? Someone who knows how to clone this thing. A method on Foo ?

trait Foo {
    fn value(&self) -> i32;
    fn clone_dyn(&self) -> Box<dyn Foo>;
}

impl Foo for Bar {
    fn value(&self) -> i32 {
        self.x
    }
    
    fn clone_dyn(&self) -> Box<dyn Foo> {
        Box::new(self.clone()) // Forward to the derive(Clone) impl
    }
}

NOW!

impl Clone for Box<dyn Foo> {
    fn clone(&self) -> Self {
        self.clone_dyn()
    }
}

IT WORKS!!

fn sort_and_print(v: &Vec<Box<dyn Foo>>) {
    let mut v_copy = v.clone();
    v_copy.sort_by_key(|o| o.value());
    for val in v_copy {
        println!("{}", val.value());
    }
}

https://play.rust-lang.org/?version=stable&mode=debug&edition=2021&gist=d6e871711146bc3f34d9710211b4a1dd

Note: The dyn-clone crate from @dtonlay generalizes this idea.

Answer 2

You can make sort_and_print() shorter using Iterator::collect() :

fn sort_and_print(v: &[Box<dyn Foo>]) {
    let mut v_copy: Vec<_> = v.iter().collect();
    v_copy.sort_by_key(|o| o.value());
    for val in v_copy {
        println!("{}", val.value());
    }
}

Playground

As an aside, accepting a vector by reference is usually better expressed as accepting a slice, as explained here , so the above answer accepts a slice.

You can make it even shorter by using the sorted() method from the itertools crate:

use itertools::Itertools;

fn sort_and_print(v: &[Box<dyn Foo>]) {
    for val in v.iter().sorted_by_key(|o| o.value()) {
        println!("{}", val.value());
    }
}

You almost certainly don't want to clone the vector because it would involve a deep copy, ie cloning each Box<dyn Foo> , which is unnecessary, potentially expensive, as well as complicated (as explained in the other answer).