“借入的价值不能活得足够长"使用构建器模式时 [英] "borrowed value does not live long enough" when using the builder pattern

问题描述

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:

I would like to be able to do this:

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

我收到此错误:

错误:借入的价值寿命不足 let mut canvas = Canvas :: new(100，100，& renderer)

error: borrowed value does not live long enough let mut canvas = Canvas::new(100, 100, &renderer)

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

Why does the returned Canvas object not live long enough? If I store the result in an intermediate let, then it works; why?

推荐答案

以下是最小复制:

#[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);
}

2015年锈蚀

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

2018年锈蚀

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)，该方法将返回引用.在方法链的最后，您试图返回引用并将其存储在变量中，但是引用指向的是无处居住的临时项！ Rust不允许您引用无效的内容.

---

The problem arises because you create a temporary variable (1), then pass the reference to that variable to the method (2), which returns the reference. At the end of the method chain, you are trying to return the reference and store it in the variable, but the reference points to a temporary item that has nowhere to live! Rust doesn't let you have a reference to something that's invalid.

部分问题是该不是构建器模式，这只是使用链式方法调用对其自身进行修改的结构.一些解决方案:

Part of the problem is that this is not the Builder pattern, this is just a struct that modifies itself using chained method invocation. Some solutions:

1. 存储临时"变量，在这种情况下，所有方法都是随后发生的普通突变方法.
2. 接受self而不是对self(&self，&mut self)的引用，然后最终返回完整的结构.
3. 在链的末尾具有build方法，该方法返回另一个独立的结构，而不是引用.

1. Store the "temporary" variable, in which case all the methods are just normal mutation methods that happen afterward.
2. Take in self instead of a reference to self (&self, &mut self) and then ultimately return the full struct.
3. Have a build method at the end of the chain that returns another standalone struct, not a reference.

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