rust - 如何克隆 struct include `Rc` ？

Question

我想定义一个类型，包括Rc<Fn(T)> , T 不需要 Clone特征，示例代码:

use std::rc::Rc;


struct X;

#[derive(Clone)]
struct Test<T> {
    a: Rc<Fn(T)>
}


fn main() {
    let t: Test<X> = Test {
        a: Rc::new(|x| {})
    };
    let a = t.clone();
}

无法编译，错误信息是:

test.rs:16:15: 16:22 note: the method `clone` exists but the following trait bounds were not satisfied: `X : core::clone::Clone`, `X : core::clone::Clone`
test.rs:16:15: 16:22 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:
test.rs:16:15: 16:22 help: candidate #1: `core::clone::Clone`
error: aborting due to previous error

如何更正我的代码？

Answer 1

问题是 #[derive(Clone)] 相当愚蠢。作为其扩展的一部分，它向所有 泛型类型参数添加了一个Clone 约束，无论它是否真的需要这样的约束。

因此，您需要手动实现Clone，如下所示:

struct Test<T> {
    a: Rc<Fn(T)>
}

impl<T> Clone for Test<T> {
    fn clone(&self) -> Self {
        Test {
            a: self.a.clone(),
        }
    }
}