派生特征会导致意外的编译器错误，但手动实现有效

Question

此代码（playground）：

#[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[E0599]: no method named `clone` found for struct `Foo<'a, T>` in the current scope
   --> src/main.rs:16:9
    |
3   | struct Foo<'a, T: 'a> {
    | ---------------------
    | |
    | method `clone` not found for this
    | doesn't satisfy `Foo<'_, T>: std::clone::Clone`
...
16  |     foo.clone();
    |         ^^^^^ method not found in `Foo<'a, T>`
    |
    = note: the method `clone` exists but the following trait bounds were not satisfied:
            `T: std::clone::Clone`
            which is required by `Foo<'_, T>: std::clone::Clone`
help: consider restricting the type parameter to satisfy the trait bound
    |
3   | struct Foo<'a, T: 'a> where T: std::clone::Clone {
    |                       ^^^^^^^^^^^^^^^^^^^^^^^^^^

添加use std::clone::Clone; 不会改变任何东西，因为它已经在前奏中了。

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

impl<'a, T> Clone for Foo<'a, T> {
    fn clone(&self) -> Self {
        Foo {
            t: self.t,
        }
    }
}

这是怎么回事？

• #[derive()]-impls 和手动的有区别吗？
• 这是编译器错误吗？
• 还有什么我没想到的？

Answer 1

答案隐藏在错误信息中：

    = note: the method `clone` exists but the following trait bounds were not satisfied:
            `T: std::clone::Clone`
            which is required by `Foo<'_, T>: std::clone::Clone`

当您派生 Clone（以及许多其他自动派生的类型）时，它会在 all 泛型类型上添加 Clone 绑定。使用rustc -Z unstable-options --pretty=expanded，我们可以看到它变成了什么：

impl <'a, T: ::std::clone::Clone + 'a> ::std::clone::Clone for Foo<'a, T> {
    #[inline]
    fn clone(&self) -> Foo<'a, T> {
        match *self {
            Foo { t: ref __self_0_0 } =>
            Foo{t: ::std::clone::Clone::clone(&(*__self_0_0)),},
        }
    }
}

在 this 的情况下，不需要绑定，因为泛型类型在引用后面。

现在，您需要自己实现Clone。 There's a Rust issue for this，但这是一个相对罕见的情况，有解决方法。

Answer 2

如果您明确标记T 应该实现Clone，您的示例将毫无问题地派生Clone，如下所示：

#[derive(Clone)]
struct Foo<'a, T: 'a> {
    t: &'a T,
}

fn bar<'a, T: Clone>(foo: Foo<'a, T>) {
    foo.clone();
}

(Playground link)

您可以避免显式指定边界似乎不寻常，但 Shepmaster 的回答似乎暗示编译器会隐式插入它，所以我的建议在功能上是相同的。