什么是特性`core :: types :: Sized`没有实现生锈中的类型`<generic#0>`？

Question

我希望这可行:

trait Task<R, E> {
  fn run(&self) -> Result<R, E>;
}

mod test {

  use super::Task;

  struct Foo;
  impl<uint, uint> Task<uint, uint> for Foo {
    fn run(&self) -> Result<uint, uint> {
      return Err(0);
    }
  }

  fn can_have_task_trait() {
    Foo;
  }
}

fn main() {
  test::can_have_task_trait();
}

......但它没有:

<anon>:10:3: 14:4 error: the trait `core::kinds::Sized` is not implemented for the type `<generic #0>`
<anon>:10   impl<uint, uint> Task<uint, uint> for Foo {
<anon>:11     fn run(&self) -> Result<uint, uint> {
<anon>:12       return Err(0);
<anon>:13     }
<anon>:14   }
<anon>:10:3: 14:4 note: the trait `core::kinds::Sized` must be implemented because it is required by `Task`
<anon>:10   impl<uint, uint> Task<uint, uint> for Foo {
<anon>:11     fn run(&self) -> Result<uint, uint> {
<anon>:12       return Err(0);
<anon>:13     }
<anon>:14   }
<anon>:10:3: 14:4 error: the trait `core::kinds::Sized` is not implemented for the type `<generic #1>`
<anon>:10   impl<uint, uint> Task<uint, uint> for Foo {
<anon>:11     fn run(&self) -> Result<uint, uint> {
<anon>:12       return Err(0);
<anon>:13     }
<anon>:14   }
<anon>:10:3: 14:4 note: the trait `core::kinds::Sized` must be implemented because it is required by `Task`
<anon>:10   impl<uint, uint> Task<uint, uint> for Foo {
<anon>:11     fn run(&self) -> Result<uint, uint> {
<anon>:12       return Err(0);
<anon>:13     }
<anon>:14   }
error: aborting due to 2 previous errors
playpen: application terminated with error code 101
Program ended.

围栏:http://is.gd/kxDt0P

发生什么了？

我不知道这个错误意味着什么.

是我正在使用Result并且要求U,V不是大小的吗？在哪种情况下,为什么它们大小？我没写:

Task<Sized? R, Sized? E>

现在所有仿制药都是动态尺寸还是什么？(在这种情况下,什么是大小？甚至是什么意思？)

这是怎么回事？

Answer 1

你只需要删除<uint, uint>in impl<uint, uint>,因为类型参数去那里而不是具体类型:

impl Task<uint, uint> for Foo { ... }

我认为你得到的错误是编译器对未使用的类型参数感到困惑.它也出现在这个超级缩减版本中:

trait Tr {}
impl<X> Tr for () {}