Clone value of reference variable Not working

help
1 
#![allow(warnings)]

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


struct Example<T>(&'static str, T);

impl<T> Deref for Example<T> {
    type Target = T;
    fn deref(&self) -> &Self::Target {
        &self.1
    }
}

impl<T> DerefMut for Example<T> {
    fn deref_mut(&mut self) -> &mut Self::Target {
        &mut self.1
    }
}

impl<T> Example<T> {
    fn get(&self) -> T {
        self.1
    }
}

fn main() {
    let mut example = Example::<String>("Albert", String::from("000-000"));
    println!("{}", example.get());
}

(Playground)

move occurs because `self.1` has type `T`, which does not implement the `Copy` trait

How can I get value T from get() function? It can't return &T, should return T

2

In general case there's no way.

If you can restrict types on which get can be called, you'll be able to clone explicitly:

impl<T> Example<T> {
    fn get(&self) -> T where T: Clone {
        self.1.clone()
    }
}
3

I can restrict types for struct Example or implementation.
But in case the restricted types don't have clone() function, is there other way to get values?

4

No. If some type doesn't implement Clone, it is single-ownership only. So, while your Example<T> is still alive, you can't get T out of it.
There are two possible workarounds, but they both change semantics of the code:

  • Store an Option<T> inside Example<T> and call Option::take inside get.
  • Make get take self instead of &self.
5

Sounds good! If possible ,can you write an example in playground and share it with me?
Thanks

6

Here's 3 examples in the playground, each of them with pros and cons Rust Playground

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7

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