无法调用引用实现特征的类型的函数

Question

I am having some trouble to understand how to work with Traits and ownerships. The following example works:

struct X([u8; 4]);

impl X {
   pub fn get(&self, n: usize) -> u8 {
       self.0[n]
   }
}

fn f1(x: &X) {
    println!("{}", x.get(1));
    f2(&x);
}

fn f2(x: &X) {
    println!("{}", x.get(2));    
}

fn main() {
    let z1 = X([1u8, 2u8, 3u8, 4u8]);
    f1(&z1);
}

But when I try to create a trait (here XT ) with get :

trait XT {
  fn get(&self, n: usize) -> u8;
}

struct X([u8; 4]);

impl XT for X {
   fn get(&self, n: usize) -> u8 {
       self.0[n]
   }
}

fn f1<T: XT>(x: &T) {
    println!("{}", x.get(1));
    f2(&x);
}

fn f2<T: XT>(x: &T) {
    println!("{}", x.get(2));    
}

fn main() {
    let z1 = X([1u8, 2u8, 3u8, 4u8]);
    f1(&z1);
}

Fails to compile with the following error message:

the trait XT is not implemented for the type &T

It works if I change f2(&x) to f2(x) . My expectation was that replacing the types by the traits, everything will work.

Answer 1

The problem is that you're trying to pass &&T to f2 . That means it expects &T to implement XT , and that's not what you said: you said that T implements XT .

You can modify f1 to properly express this constraint by using a where T: XT, for<'a> &'a T: XT clause, but then you can't call f1 from main because &X doesn't implement XT . So you go and add an implementation for that as well , and then the code works... but honestly, it's easier to just remove that & and call f2(x) instead.

To put it another way: just because a type implements a trait does not mean pointers to that type also implement the trait.