为什么使用 "Self" 作为参数类型会引发生命周期错误？

Question

我目前正在关注 https://raytracing.github.io/books/RayTracingInOneWeekend.html，但我正在用 Rust 实现所有内容。这是我的向量实现的摘录：

type Scalar = f64;

#[derive(Debug, Default, Clone)]
pub struct Vector {
    x: Scalar,
    y: Scalar,
    z: Scalar,
}

impl Vector {
    fn new(x: Scalar, y: Scalar, z: Scalar) -> Self {
        Self { x, y, z }
    }

    fn x(&self) -> Scalar {
        self.x
    }

    fn y(&self) -> Scalar {
        self.y
    }

    fn z(&self) -> Scalar {
        self.z
    }
}

impl std::ops::Mul<&Vector> for &Vector {
    type Output = Scalar;

    fn mul(self, rhs: Self) -> Self::Output {
        self.x() * rhs.x() + self.y() * rhs.y() + self.z() * rhs.z()
    }
}

当我尝试编译它时，我收到以下消息：

error[E0308]: method not compatible with trait
  --> src/point.rs:33:5
   |
33 |     fn mul(self, rhs: Self) -> Self::Output {
   |     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ lifetime mismatch
   |
   = note: expected fn pointer `fn(&point::Vector, &point::Vector) -> _`
              found fn pointer `fn(&point::Vector, &point::Vector) -> _`
note: the lifetime `'_` as defined on the impl at 30:20...
  --> src/point.rs:30:20
   |
30 | impl std::ops::Mul<&Vector> for &Vector {
   |                    ^
note: ...does not necessarily outlive the lifetime `'_` as defined on the impl at 30:20
  --> src/point.rs:30:20
   |
30 | impl std::ops::Mul<&Vector> for &Vector {
   |                    ^

error: aborting due to previous error

For more information about this error, try `rustc --explain E0308`.
error: could not compile `raytracing`.

但是，如果我将 mul 函数上的 Self 参数更改为 &Vector，它编译得很好：

[...]
fn mul(self, rhs: &Vector) -> Self::Output {
[...]

这只是生命周期推断失败的情况吗？如果是这样，为什么会失败，因为编译器似乎已正确推断出所有内容？

Answer 1

因为lifetime elision的规则，报错信息告诉它：

note: the lifetime '_ as defined on the impl at 30:20...

行：

impl std::ops::Mul<&Vector> for &Vector {

解释为：

impl<'a, 'b> std::ops::Mul<&'a Vector> for &'b Vector // Self is &'b Vector

因此生命周期不匹配，因为：

fn mul(self, rhs: Self) -> Self::Output {

是

fn mul(self, rhs: &'b Vector) -> Self::Output {

&'a Vector != &'b Vector 所以它无法编译。因为 rhs 应该是 &'a Vector.

当您使用Self时：

impl std::ops::Mul<Self> for &Vector {

成为：

impl<'a> std::ops::Mul<&'a Vector> for &'a Vector {

所以在 fn mul(self, rhs: Self) -> Self::Output { rhs 中会有正确的生命周期 <'a>

如果出现生命周期问题，请尝试明确检查编译器是否出错。

最终代码不应包含任何 Self 关键字以允许不同的生命周期：

impl std::ops::Mul<&Vector> for &Vector {
    type Output = Scalar;

    fn mul(self, rhs: &Vector) -> Self::Output {
        self.x() * rhs.x() + self.y() * rhs.y() + self.z() * rhs.z()
    }
}

显式：

impl<'a, 'b> std::ops::Mul<&'a Vector> for &'b Vector {
    type Output = Scalar;

    fn mul(self, rhs: &'a Vector) -> Self::Output {
        self.x() * rhs.x() + self.y() * rhs.y() + self.z() * rhs.z()
    }
}

为什么使用 "Self" 作为参数类型会引发生命周期错误？

Why does using "Self" as a parameter type raise a lifetime error?

type-inference

lifetime

rust