10

我有以下代码：

pub struct Canvas<'a> {
    width: isize,
    height: isize,
    color: Color,
    surface: Surface,
    texture: Texture,
    renderer: &'a Renderer,
}

impl<'a> Canvas<'a> {
    pub fn new(width: isize, height: isize, renderer: &'a Renderer) -> Canvas<'a> {
        let color = Color::RGB(0, 30, 0);
        let mut surface = core::create_surface(width, height);
        let texture = Canvas::gen_texture(&mut surface, width, height, color, renderer);
        Canvas {
            width: width,
            height: height,
            color: color,
            surface: surface,
            texture: texture,
            renderer: renderer,
        }
    }

    pub fn color(&mut self, color: Color) -> &mut Canvas<'a> {
        self.color = color;
        self.texture = Canvas::gen_texture(
            &mut self.surface,
            self.width,
            self.height,
            self.color,
            self.renderer,
        );
        self
    }
}

我希望能够做到这一点：

let mut canvas = Canvas::new(100, 100, &renderer).color(Color::RGB(80, 230, 80));

我收到此错误：

错误：借来的值不够长让 mut canvas = Canvas::new(100, 100, &renderer)

为什么返回的Canvas对象寿命不够长？如果我将结果存储在中间let，那么它可以工作；为什么？

4

1 回答 1

17

这是一个最小的复制：

#[derive(Debug)]
pub struct Canvas;

impl Canvas {
    fn new() -> Self {
        Canvas
    }

    fn color(&self) -> &Canvas {
        self
    }
}

fn main() {
    let mut canvas = Canvas::new().color();
    //             1 ^~~~~~~~~~~~~
    //                           2 ^~~~~
    println!("{:?}", canvas);
}

生锈 2015

error[E0597]: borrowed value does not live long enough
  --> src/main.rs:15:22
   |
15 |     let mut canvas = Canvas::new().color();
   |                      ^^^^^^^^^^^^^        - temporary value dropped here while still borrowed
   |                      |
   |                      temporary value does not live long enough
...
19 | }
   | - temporary value needs to live until here
   |
   = note: consider using a `let` binding to increase its lifetime

生锈 2018

error[E0716]: temporary value dropped while borrowed
  --> src/main.rs:15:22
   |
15 |     let mut canvas = Canvas::new().color();
   |                      ^^^^^^^^^^^^^        - temporary value is freed at the end of this statement
   |                      |
   |                      creates a temporary which is freed while still in use
...
18 |     println!("{:?}", canvas);
   |                      ------ borrow later used here
   |
   = note: consider using a `let` binding to create a longer lived value

出现问题是因为您创建了一个临时变量 (1)，然后将对该变量的引用传递给方法 (2)，该方法返回引用。在方法链的末尾，您试图返回引用并将其存储在变量中，但引用指向一个无处可存的临时项目！Rust 不允许你引用无效的东西。

部分问题在于这不是 Builder 模式，这只是一个使用链式方法调用修改自身的结构。一些解决方案：

  1. 存储“临时”变量，在这种情况下，所有方法都只是之后发生的正常突变方法。
  2. 代替self对 self ( &self, &mut self) 的引用，然后最终返回完整的结构。
  3. build在链的末尾有一个方法，该方法返回另一个独立的结构，而不是引用。
于 2015-02-12T05:16:06.490 回答