CODEWITHSUNDEEP
rusttype-conversion
K. Biermann
K. Biermann

Reputation: 1319

How to overload the reference operator?

I've implemented AsRef<[u8]> for my type and the automatic conversion to &[u8] using .as_ref() works but the &-operator doesn't... how can I make the operator work?

struct Buffer {
    data: Vec<u8>,
}

impl AsRef<[u8]> for Buffer {
    fn as_ref(&self) -> &[u8] {
        &self.data
    }
}

fn print_slice(slice: &[u8]) {
    slice.iter().for_each(|b| print!("{:02x}", b));
    println!()
}

fn main() {
    let buffer = Buffer {
        data: b"Testolope".to_vec(),
    };
    // print_slice(&buffer);     // <- This does not work
    print_slice(buffer.as_ref()) // <- This works
}

error[E0308]: mismatched types
  --> src/main.rs:20:17
   |
20 |     print_slice(&buffer);
   |                 ^^^^^^^ expected slice, found struct `Buffer`
   |
   = note: expected type `&[u8]`
              found type `&Buffer`

I want a generic solution. Other datatypes like Vec<u8> support the conversion to &[u8] by using the &-operator. It would be cool if I could make this work for my own types so that I don't have to use .as_ref() every time.

Upvotes: 2

Views: 822

Answers (2)

Shepmaster
Shepmaster

Reputation: 430634

You cannot overload the reference operator, but you can hook into Deref coercion:

impl std::ops::Deref for Buffer {
    type Target = [u8];

    fn deref(&self) -> &Self::Target {
        &self.data
    }
}

Your original code works in this case.

See also:

Upvotes: 4

stevensonmt
stevensonmt

Reputation: 732

You are calling the whole struct with print_slice(&buffer); but only the data field with print_slice(buffer.as_ref()). If you make it print_slice(&buffer.data) it will work. Alternatively change the type signature of the print_slice function to expect &Buffer, which will make the as_ref() line not work.

Upvotes: 1

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