如何实现“添加"特征来引用结构? [英] How do I implement the Add trait for a reference to a struct?

问题描述

我制作了两个元素Vector结构，并且想重载+运算符.

I made a two element Vector struct and I want to overload the + operator.

我使所有函数和方法都采用引用而不是值，而且我希望+运算符以相同的方式工作.

I made all my functions and methods take references, rather than values, and I want the + operator to work the same way.

impl Add for Vector {
    fn add(&self, other: &Vector) -> Vector {
        Vector {
            x: self.x + other.x,
            y: self.y + other.y,
        }
    }
}

根据我尝试的变化，我会遇到生命周期问题或类型不匹配.具体来说，&self参数似乎没有被视为正确的类型.

Depending on which variation I try, I either get lifetime problems or type mismatches. Specifically, the &self argument seems to not get treated as the right type.

我在impl和Add上都看到了带有模板参数的示例，但是它们只会导致不同的错误.

I have seen examples with template arguments on impl as well as Add, but they just result in different errors.

我发现如何为不同的RHS类型和返回值重载运算符? /a>，但是即使我将use std::ops::Mul;放在顶部，答案中的代码也无法正常工作.

I found How can an operator be overloaded for different RHS types and return values? but the code in the answer doesn't work even if I put a use std::ops::Mul; at the top.

我每晚使用rustc 1.0.0(ed530d7a3 2015-01-16 22:41:16 +0000)

I am using rustc 1.0.0-nightly (ed530d7a3 2015-01-16 22:41:16 +0000)

我不会接受您只有两个字段，为什么要使用引用"作为答案；如果我想要一个100个元素的结构怎么办?我将接受一个答案，该答案表明即使是大型结构，如果是这种情况，我也应该按值传递(尽管我不这么认为.)我有兴趣了解结构大小的良好经验法则并按值和结构传递值，但这不是当前的问题.

I won't accept "you only have two fields, why use a reference" as an answer; what if I wanted a 100 element struct? I will accept an answer that demonstrates that even with a large struct I should be passing by value, if that is the case (I don't think it is, though.) I am interested in knowing a good rule of thumb for struct size and passing by value vs struct, but that is not the current question.

推荐答案

您需要在&Vector而不是Vector上实现Add.

You need to implement Add on &Vector rather than on Vector.

impl<'a, 'b> Add<&'b Vector> for &'a Vector {
    type Output = Vector;

    fn add(self, other: &'b Vector) -> Vector {
        Vector {
            x: self.x + other.x,
            y: self.y + other.y,
        }
    }
}

在其定义中，Add::add始终将self作为值.但是引用的类型与其他任何 ¹ 一样，因此它们也可以实现特征.在引用类型上实现特征时，self的类型是引用；引用按值传递.通常，Rust中按值传递意味着转移所有权，但是当按值传递引用时，它们只是被复制(如果是可变引用，则被重新借入/移动)，并且不会转移对引用者的所有权(因为引用首先不拥有其参照对象).考虑到所有这些，Add::add(和许多其他运算符)按值取self是有意义的:如果需要获取操作数的所有权，则可以直接在结构/枚举上实现Add，如果否，您可以在引用上实现Add.

In its definition, Add::add always takes self by value. But references are types like any other¹, so they can implement traits too. When a trait is implemented on a reference type, the type of self is a reference; the reference is passed by value. Normally, passing by value in Rust implies transferring ownership, but when references are passed by value, they're simply copied (or reborrowed/moved if it's a mutable reference), and that doesn't transfer ownership of the referent (because a reference doesn't own its referent in the first place). Considering all this, it makes sense for Add::add (and many other operators) to take self by value: if you need to take ownership of the operands, you can implement Add on structs/enums directly, and if you don't, you can implement Add on references.

在这里，self的类型为&'a Vector，因为这是我们要在其上实现Add的类型.

Here, self is of type &'a Vector, because that's the type we're implementing Add on.

请注意，我还为RHS类型参数指定了不同的生存期，以强调两个输入参数的生存期无关的事实.

Note that I also specified the RHS type parameter with a different lifetime to emphasize the fact that the lifetimes of the two input parameters are unrelated.

¹ 实际上，引用类型的特殊之处在于，您可以为包装箱中定义的类型实现对引用的特征(即，如果允许您为T实现特征，那么您可以还允许将其用于&T). &mut T和Box<T>具有相同的行为，但通常对于U<T>而言并非如此，其中U不在同一包装箱中定义.

¹ Actually, reference types are special in that you can implement traits for references to types defined in your crate (i.e. if you're allowed to implement a trait for T, then you're also allowed to implement it for &T). &mut T and Box<T> have the same behavior, but that's not true in general for U<T> where U is not defined in the same crate.

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