How do I implement a trait I don't own for a type I don't own?

Question

I wanted to implement the Shl trait for Vec, the code is below. This would make things like vec << 4 possible, which would be ni

Answer 1

This would make things like vec << 4 possible, which would be nice sugar for vec.push(4).

Although it can be done, it is generally a bad idea to implement an operator with a unexpected semantic.

Here is an example of how this can be done:

use std::ops::Shl;

struct BadVec<T>(Vec<T>);

impl<T> Shl<T> for BadVec<T> {
    type Output = BadVec<T>;

    fn shl(mut self, elem: T) -> Self::Output {
        self.0.push(elem);
        self
    }
}

fn main() {
    let mut v = BadVec(vec![1, 2, 3]);
    v = v << 4;
    assert_eq!(vec![1, 2, 3, 4], v.0)
}

If you implement Deref (DerefMut):

use std::ops::{Deref, DerefMut};

impl<T> Deref for BadVec<T> {
    type Target = Vec<T>;

    fn deref(&self) -> &Self::Target {
        &self.0
    }
}

impl<T> DerefMut for BadVec<T> {
    fn deref_mut(&mut self) -> &mut Self::Target {
        &mut self.0
    }
}

you can call Vec methods:

fn main() {
    let mut v = BadVec(vec![1, 2, 3]);
    v = v << 4;
    v.truncate(2);
    assert_eq!(2, v.len());
}

---

Take a look at the newtype_derive crate, it can generate some boilerplate code for you.

Answer 2

While you can't do that exactly, the usual workaround is to just wrap the type you want in your own type and implement the trait on that.

use somecrate::FooType;
use somecrate::BarTrait;

struct MyType(FooType);

impl BarTrait for MyType {
    fn bar(&self) {
        // use `self.0` here
    }
}