在Rust中模仿克隆特征

Question

我有一个简单的链表类型和它的实现Clone:

#[deriving(Show)]
enum List {
    Cons(int, Box<List>),
    Nil,
}

impl Clone for List {
    fn clone(&self) -> List {
        match *self {
            Cons(val, ref rest) => Cons(val, rest.clone()),
            Nil => Nil,
        }
    }
}

它按预期工作.但是,如果我做我自己的MyClone特质与 相同签名 作为的Clone,我得到一个错误:

trait MyClone {
    fn my_clone(&self) -> Self;
}

impl MyClone for List {
    fn my_clone(&self) -> List {
        match *self {
            Cons(val, ref rest) => Cons(val, rest.my_clone()),
            Nil => Nil,
        }
    }
}

.../src/main.rs:23:46: 23:61 error: mismatched types: expected `Box<List>`, found `List` (expected box, found enum List)
.../src/main.rs:23             Cons(val, ref rest) => Cons(val, rest.my_clone()),

如果我改变它box rest.my_clone(),它工作正常,但我不明白为什么.在MyClone和Clone特点是相同的,所以在我看来,他们将接受相同的实现.

(我正在用rustc 0.12.0-nightly编译(72841b128 2014-09-21 20:00:29 +0000).)

Answer 1

这是因为锈也有克隆的实现Box<T>.如果你实现MyClone它,它将是预期的owrk.

#[deriving(Show)]
enum List {
    Cons(int, Box<List>),
    Nil,
}

trait MyClone {
    fn my_clone(&self) -> Self;
}

impl<T: MyClone> MyClone for Box<T> {
    fn my_clone(&self) -> Box<T> {
        self.my_clone()
    }
}

impl MyClone for List {
    fn my_clone(&self) -> List {
        match *self {
            Cons(val, ref rest) => Cons(val, rest.my_clone()),
            Nil => Nil,
        }
    }
}