派生特征会导致意外的编译器错误,但手动实现会起作用

Question

这段代码(游乐场):

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

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

......不编译:

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;不会改变任何东西,因为它已经在前奏中了.

当我删除#[derive(Clone)]并手动执行Clone时Foo,它按预期编译!

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manual 之间有区别吗？
• 这是编译器错误吗？
• 我还没有想到的其他东西？

Answer 1

答案隐藏在错误消息中:

该方法clone存在,但不满足以下特征界限:T : std::clone::Clone

当您派生Clone(以及许多其他自动派生类型)时,它会Clone在所有泛型类型上添加一个绑定.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)),},
        }
    }
}

在这种情况下,不需要绑定,因为泛型类型在引用后面.

现在,您需要Clone自己实施.这有一个Rust问题,但这是一个相对罕见的解决方法.