copy - 按值重载运算符会导致使用移动的值

Question

编译以下使用运算符重载的 Rust 代码

use std::ops::{Add};

#[derive(Show)]
struct Point {
    x: int,
    y: int
}

impl Add for Point {
    type Output = Point;

    fn add(self, other: Point) -> Point {
        Point {x: self.x + other.x, y: self.y + other.y}
    }
}

fn main() {
    let p: Point = Point {x: 1, y: 0};
    let pp = p + p;
}

由于 p 的所有权导致编译器错误:

<anon>:21:18: 21:19 error: use of moved value: `p`
<anon>:21     let pp = p + p;
                           ^
<anon>:21:14: 21:15 note: `p` moved here because it has type `Point`, which is non-copyable
<anon>:21     let pp = p + p;
                       ^

解释其背后的基本原理here并导致RFC未被接受(部分原因是上述示例)。然而，后来以下RFC仍然为运算符引入了按值类型签名。

虽然我理解该决定背后的理由。由于我缺乏使用rust 经验，我不确定允许上述代码工作的“正确”方法是什么(a)如果我不想复制或(b)如何使结构可复制？

Answer 1

如果您不想复制，就我的新手理解而言，您需要在对 Point 的引用上实现 Add。

这将得到 RFC 的支持:

Fortunately, there is no loss in expressiveness, since you can always implement the trait on reference types. However, for types that do need to be taken by reference, there is a slight loss in ergonomics since you may need to explicitly borrow the operands with &. The upside is that the ownership semantics become clearer: they more closely resemble normal function arguments.

确实它似乎有效:

use std::ops::{Add};

#[derive(Show)]
struct Point {
    x: i32,
    y: i32
}

impl<'a> Add for &'a Point {
    type Output = Point;

    fn add(self, other: &'a Point) -> Point { //'
        Point {x: self.x + other.x, y: self.y + other.y}
    }
}

fn main() {
    let p: Point = Point {x: 1, y: 0};
    let pp = &p + &p;
    println!("{:?}", pp);
}

( playpen )

要使 Point 可复制，只需将 #[derive(Show)] 替换为 #[derive(Show,Copy)]。这些结构过去默认是可复制的，但它 changed .