Mut reference clone

help
1

What is the expected behavior of calling clone() on &mut T? Is it defined as calling T.clone() ? I tried in Rust playground, it seems calling T.clone(), but the doc seems to say something different.

playground test:

fn dup(v: &mut Vec<u8>) -> Vec<u8> {
    v.clone()
}

fn consume(v: Vec<u8>) {
    println!("consuming v1: {:?}", v);
}

fn main() {
    let mut v1 = vec![1, 2];
    let v2 = dup(&mut v1);  
    consume(v1);
    println!("v2: {:?}", v2);
}

(Playground)

2

It's the type coercion which allows to call methods that takes &T on &mut T.

1 Like
3

Does it mean that clone() call works same with &mut T, &T and T ?
If that's the case, the following statements in the doc are confusing:

The following traits are implemented for all &T , regardless of the type of its referent:

  • Clone (Note that this will not defer to T 's Clone implementation if it exists!)

&mut T references get all of the above except Copy and Clone (to prevent creating multiple simultaneous mutable borrows),

4

I would’ve thought it is actually automatic dereferencing.

According to the reference when a &mut Vec<u8> is encoutered, the receiver types &mut Vec<u8>, &&mut Vec<u8>, &mut &mut Vec<u8>, Vec<u8>, &Vec<u8>, &mut Vec<u8>, [T], &[T], &mut [T] are checked in this order, because of the Deref implementations of &mut T and Vec.

Then the first one of those types is used that can offer a clone method. More precisely, when we want to consider clone from Clone, that one takes its argument by-reference, so to successfully apply the Clone trait, the receiver type must be of the form &T for some T: Clone.

Let’s start trying:

  • &mut Vec<u8> is not of the form &T.
  • &&mut Vec<u8> is of the form &T for T = &mut Vec<u8>, but &mut Vec<u8> does not implement Clone.
  • &mut &mut Vec<u8> is not of the form &T.
  • Vec<u8> is not of the form &T.
  • &Vec<u8> is of the form &T for T = Vec<u8>, and Vec<u8> does implement clone

So the call v.clone() for v: &mut Vec<u8> becomes <Vec<u8>>::clone(<&mut Vec<u8>>::deref(&v))

For a sanity check, this does compile:

fn dup(v: &mut Vec<u8>) -> Vec<u8> {
    use std::ops::Deref;
    <Vec<u8>>::clone(<&mut Vec<u8>>::deref(&v))
}
3 Likes
5

Yeah, you can try to make sense of this by following my above explanation (previous comment). Perhaps you can try to apply the same logic to find out why in clone_ref1 and clone_ref2 you need to add the & to make it compile.

fn clone_ref<T>(x: &T) -> &T {
    x.clone()
}
fn clone_ref1(x: &u8) -> &u8 {
    // x.clone() does not compile here
    (&x).clone()
}
fn clone_ref2<T: Clone>(x: &T) -> &T {
    // x.clone() does not compile here
    (&x).clone()
}
1 Like
6

The dot operator is a bit magical

http://stackoverflow.com/questions/28519997/what-are-rusts-exact-auto-dereferencing-rules/28552082#28552082

7

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