派生特征会导致意外的编译器错误，但是手动实现有效

Question

此代码（操场）：

#[derive(Clone)]

struct Foo<'a, T: 'a> {

    t: &'a T,

}

fn bar<'a, T>(foo: Foo<'a, T>) {

    foo.clone();

}

...无法编译：

error: no method named `clone` found for type `Foo<'a, T>` in the current scope

  --> <anon>:7:9

   |>

16 |>     foo.clone();

   |>         ^^^^^

note: the method `clone` exists but the following trait bounds were not satisfied: `T : std::clone::Clone`

help: items from traits can only be used if the trait is implemented and in scope; the following trait defines an item `clone`, perhaps you need to implement it:

help: candidate #1: `std::clone::Clone`

添加use std::clone::Clone;并不会改变任何内容，因为它已经在序幕中了。

当我删除#[derive(Clone)]并手动实现Clone时Foo，它会按预期编译！

impl<'a, T> Clone for Foo<'a, T> {

    fn clone(&self) -> Self {

        Foo {

            t: self.t,

        }

    }

}

这里发生了什么？

#[derive()]-impls和手动的之间有区别吗？

这是编译器错误吗？

还有我没想到的东西？

Answer 1

答案隐藏在错误消息中：该方法clone存在，但不满足以下特征范围：T : std::clone::Clone当您派生Clone（以及许多其他自动派生的类型）时，它会Clone在所有通用类型上添加一个界限。使用rustc -Z unstable-options --pretty=expanded，我们可以看到它变为：impl <'a, T: ::std::clone::Clone + 'a> ::std::clone::Clone for Foo<'a, T> {&nbsp; &nbsp; #[inline]&nbsp; &nbsp; fn clone(&self) -> Foo<'a, T> {&nbsp; &nbsp; &nbsp; &nbsp; match *self {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Foo { t: ref __self_0_0 } =>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Foo{t: ::std::clone::Clone::clone(&(*__self_0_0)),},&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }}在这种情况下，不需要绑定，因为泛型类型在引用后面。现在，您将需要实现Clone自己。有一个Rust问题，但这是一种相对罕见的解决方法。