你如何在 rust 中实现泛型特征的特定类型

我最初认为您可以这样做，因为文档(http://doc.rust-lang.org/rust.html#implementations)建议您可以:

trait Bar<T> {
  fn ex(&self) -> T;
}

struct Foo {
  y:f64
}

impl Bar<int> for Foo {
  fn ex(&self) -> int {
    return self.y.floor() as int;
  }
}

impl Bar<uint> for Foo {
  fn ex(&self) -> uint {
    if (self.y < 0.0) {
      return 0u;
    }
    return self.y.floor() as uint;
  }
}

...但这似乎不起作用.我会遇到如下错误:

error: multiple applicable methods in scope
error: expected Bar<uint>, but found Bar<int> (expected uint but found int)
error: expected Bar<int>, but found Bar<uint> (expected int but found uint)

所以我想也许Foo必须是通用的，这样才能工作，所以每个特定的Foo都有自己的Bar实现:

trait Bar<T> {
  fn ex(&self) -> T;
}

struct Foo<T> {
  y:f64
}

impl<T> Foo<T> {
  fn new<U>(value:f64) -> Foo<U> {
    return Foo { y: value } as Foo<U>;
  }
}

impl Bar<int> for Foo<int> {
  fn ex(&self) -> int {
    return self.y.floor() as int;
  }
}

impl Bar<uint> for Foo<uint> {
  fn ex(&self) -> uint {
    if (self.y < 0.0) {
      return 0u;
    }
    return self.y.floor() as uint;
  }
}

fn main() {
  let z = Foo::new::<int>(100.5);
  let q = Foo::new::<uint>(101.5);
  let i:int = z.ex();
  let j:uint = q.ex();
}

...但我的构造函数似乎不起作用:

x.rs:11:12: 11:38 error: non-scalar cast: `Foo<<generic #1>>` as `Foo<U>`
x.rs:11     return Foo { y: value } as Foo<U>;
                   ^~~~~~~~~~~~~~~~~~~~~~~~~~
error: aborting due to previous error

编辑:我也试过:

impl<T> Foo<T> {
  fn new<U>(value:f64) -> Foo<U> {
    let rtn:Foo<U> = Foo { y: value };
    return rtn;
  }
}

解决了铸造误差，但导致:

x.rs:32:11: 32:26 error: cannot determine a type for this expression: unconstrained type
x.rs:32   let z = Foo::new::<int>(100.5);
                  ^~~~~~~~~~~~~~~

我不知道那是什么意思.

你是怎么做到的？