Question

I am a bit stuck (except by duplicating my memory (see DupKeyEAB in playpen), I got this trait:

   pub trait KeyVal : Send + Sync + 'static { 
      type Key : Clone + Debug + 'static;
      fn get_key(&self) -> Self::Key;
      fn get_key_ref<'a>(&'a self) -> &'a Self::Key;
    }

My problem is that I need something like (to avoid using clone each time I access the key) :

  pub trait KeyVal : Send + Sync + 'static { 
      type Key : Clone + Debug + 'static;
      type KeyRef<'_> : Debug;
      fn get_key(&self) -> Self::Key;
      fn get_key_ref<'a>(&'a self) -> Self::KeyRef<'a>;
    }

or directly (with KeyRef as a trait)

  fn get_key_ref<'a,K : KeyRef<'a>>(&'a self) -> K;

or

   fn get_key_ref<'a>(&'a self) -> KeyRef<'a>;

All those notations are obviously pretty invalid, but it illustrate that I need to return a reference with the same lifetime as the struct implementing my trait but at the same time the reference could not simply be &'a but also a enum with the same lifetime.

That way when using get_key_ref on simple struct my KeyRef is simply &'aKey and when using get_key_ref on enum combining multiple trait implementation (see EnumAB) I could use a wrapper enum over the reference to the key : like KeyABRef<'a>.

  impl EnumAB {
     fn get_key_ok<'a>(&'a self) -> KeyABRef<'a> {
       match self {
         &EnumAB::A(ref a) => KeyABRef::A(a.get_key_ref()),
         &EnumAB::B(ref b) => KeyABRef::B(b.get_key_ref()),
       }
     }
   }

Yet I cannot include this function in my trait. I wonder if anyone got a solution for this kind of need (KeyVal need to be 'static)?

My original test code was :

use std::fmt::Debug;
use std::thread;

pub trait KeyVal : Send + Sync + 'static {
  type Key : Clone + Debug + 'static;
  fn get_key(&self) -> Self::Key;
  fn get_key_ref<'a>(&'a self) -> &'a Self::Key;
}

pub fn do_something_with_spawn<KV : KeyVal> (kv : KV) {
  thread::spawn ( move || {
    println!("{:?}", kv.get_key_ref());
  });
}
#[derive(Debug)]
pub struct StructA (usize);
#[derive(Debug)]
pub struct StructB (String);

#[derive(Debug)]
pub enum EnumAB {
  A(StructA),
  B(StructB),
}


impl KeyVal for StructA {
  type Key = usize;
  // type KeyRef<'_> = &'_ usize;
  fn get_key(&self) -> Self::Key {
    self.0.clone()
  }
  fn get_key_ref<'a>(&'a self) -> &'a Self::Key {
    &self.0
  }
}

impl KeyVal for StructB {
  type Key = String;
  // type KeyRef<'_> = &'_ String;
  fn get_key(&self) -> Self::Key {
    self.0.clone()
  }
  fn get_key_ref<'a>(&'a self) -> &'a Self::Key {
    &self.0
  }
}

#[derive(Clone,Debug)]
pub enum KeyAB {
  A(usize),
  B(String),
}

#[derive(Clone,Debug)]
pub enum KeyABRef<'a> {
  A(&'a usize),
  B(&'a String),
}

impl KeyVal for EnumAB {
  type Key = KeyAB;
  // type KeyRef<'_> = KeyABRef<'_>
  fn get_key(&self) -> Self::Key {
    match self {
      &EnumAB::A(ref a) => KeyAB::A(a.get_key()),
      &EnumAB::B(ref b) => KeyAB::B(b.get_key()),
    }
  }
  fn get_key_ref<'a>(&'a self) -> &'a Self::Key {
    panic!("cannot");
  }
}

impl EnumAB {
  fn get_key_ok<'a>(&'a self) -> KeyABRef<'a> {
    match self {
      &EnumAB::A(ref a) => KeyABRef::A(a.get_key_ref()),
      &EnumAB::B(ref b) => KeyABRef::B(b.get_key_ref()),
    }
  }
}

#[derive(Debug)]
pub struct DupKeyEAB (KeyAB, EnumAB);
impl KeyVal for DupKeyEAB {
  type Key = KeyAB;
  fn get_key(&self) -> Self::Key {
    self.0.clone()
  }
  fn get_key_ref<'a>(&'a self) -> &'a Self::Key {
    &self.0
  }
}

fn main () {
  let d2 = StructB("hello".to_string());
  let d3 = EnumAB::A(StructA(3));
  println!("{:?}",d2.get_key());
  println!("{:?}",d3.get_key());
  println!("{:?}",d2.get_key_ref());
  println!("{:?}",d3.get_key_ok());

 do_something_with_spawn(d3);
}

Answer 1

我同意A.B.你的动机不明确，所以我们提供的任何帮助都是最好的猜测。话虽如此，这是一个使用Borrow的猜测：

use std::borrow::Borrow;

trait KeyVal<'a> { 
    type Key: Borrow<Self::KeyRef>;
    type KeyRef: 'a;

    fn get_key(&self) -> Self::Key;
    fn get_key_ref(&'a self) -> &'a Self::KeyRef;
}

struct Example(u8);

impl<'a> KeyVal<'a> for Example {
    type Key = u8;
    type KeyRef = u8;

    fn get_key(&self) -> Self::Key {
        self.0
    }  

    fn get_key_ref(&'a self) -> &'a Self::KeyRef {
        &self.0
    }
}

fn main() {
    let e = Example(42);
    println!("{:?}", e.get_key());
    println!("{:p}", e.get_key_ref());
}

另一部分是type KeyRef: 'a，表示所选类型必须比'a更长。

我认为你不需要Borrow，我想：

trait KeyVal<'a> { 
    type Key;
    type KeyRef: 'a;

    fn get_key(&self) -> Self::Key;
    fn get_key_ref(&'a self) -> Self::KeyRef;
}

#[derive(Debug)]
struct Example(u8);
#[derive(Debug)]
struct RefWrapper<'a>(&'a u8);

impl<'a> KeyVal<'a> for Example {
    type Key = u8;
    type KeyRef = RefWrapper<'a>;

    fn get_key(&self) -> Self::Key {
        self.0
    }  

    fn get_key_ref(&'a self) -> Self::KeyRef {
        RefWrapper(&self.0)
    }
}

fn main() {
    let e = Example(42);
    println!("{:?}", e.get_key());
    println!("{:?}", e.get_key_ref());
}

Answer 2

使用Shepmaster的解决方案（加上产生的HRTB），它变为：

use std::fmt::Debug;

use std::thread;

pub trait KeyVal<'a> : Send + Sync + 'static {
  type Key : Clone + Debug + 'static;
  type KeyRef : Debug + 'a;
  fn get_key(&self) -> Self::Key;
  fn get_key_ref(&'a self) -> Self::KeyRef;
}

pub fn do_something_with_spawn<KV> (kv : KV)
  where for <'a> KV : KeyVal<'a> {
  thread::spawn ( move || {
  println!("{:?}", kv.get_key_ref());
  });
}
#[derive(Debug)]
pub struct StructA (usize);
#[derive(Debug)]
pub struct StructB (String);

#[derive(Debug)]
pub enum EnumAB {
  A(StructA),
  B(StructB),
}


impl<'a> KeyVal<'a> for StructA {
  type Key = usize;
  type KeyRef = &'a usize;
  fn get_key(&self) -> Self::Key {
    self.0.clone()
  }
  fn get_key_ref(&'a self) -> Self::KeyRef {
    &self.0
  }
}

impl<'a> KeyVal<'a> for StructB {
  type Key = String;
  type KeyRef = &'a String;
  fn get_key(&self) -> Self::Key {
    self.0.clone()
  }
  fn get_key_ref(&'a self) -> Self::KeyRef {
    &self.0
  }
}

#[derive(Clone,Debug)]
pub enum KeyAB {
  A(usize),
  B(String),
}

#[derive(Clone,Debug)]
pub enum KeyABRef<'a> {
  A(&'a usize),
  B(&'a String),
}

impl<'a> KeyVal<'a> for EnumAB {
  type Key = KeyAB;
  type KeyRef = KeyABRef<'a>;
  fn get_key(&self) -> Self::Key {
    match self {
      &EnumAB::A(ref a) => KeyAB::A(a.get_key()),
      &EnumAB::B(ref b) => KeyAB::B(b.get_key()),
    }
  }
  fn get_key_ref(&'a self) -> Self::KeyRef {
    match self {
      &EnumAB::A(ref a) => KeyABRef::A(a.get_key_ref()),
      &EnumAB::B(ref b) => KeyABRef::B(b.get_key_ref()),
    }
  }
}



fn main () {
  let d2 = StructB("hello".to_string());
  let d3 = EnumAB::A(StructA(3));
  println!("{:?}",d2.get_key());
  println!("{:?}",d3.get_key());
  println!("{:?}",d2.get_key_ref());
  println!("{:?}",d3.get_key_ref());
  do_something_with_spawn(d3);

}

How to return a reference from a trait when combining types into enum

2 个答案: