相关疑难解决方法(0)

我不明白这个错误cannot move out of borrowed content.我收到了很多次,我总是解决它,但我从来没有理解为什么.

例如:

for line in self.xslg_file.iter() {
    self.buffer.clear();

    for current_char in line.into_bytes().iter() {
        self.buffer.push(*current_char as char);
    }

    println!("{}", line);
}

产生错误:

error[E0507]: cannot move out of borrowed content
  --> src/main.rs:31:33
   |
31 |             for current_char in line.into_bytes().iter() {
   |                                 ^^^^ cannot move out of borrowed content

我通过克隆解决了这个问题line:

error[E0507]: cannot move out of `*line` which is behind a shared reference
  --> src/main.rs:31:33
   |
31 |             for current_char in line.into_bytes().iter() {
   |                                 ^^^^ …

在这种情况下，错误意味着什么：

fn main() {
    let mut v: Vec<usize> = vec![1, 2, 3, 4, 5];
    v[v[1]] = 999;
}

fn main() {
    let mut v: Vec<usize> = vec![1, 2, 3, 4, 5];
    v[v[1]] = 999;
}

我发现索引是通过Index和IndexMut特性实现的，这v[1]是*v.index(1). 有了这些知识，我尝试运行以下代码：

use std::ops::{Index, IndexMut};

fn main() {
    let mut v: Vec<usize> = vec![1, 2, 3, 4, 5];
    *v.index_mut(*v.index(1)) = 999;
}

令我惊讶的是，这完美无缺！为什么第一个片段不起作用，但第二个片段起作用？我理解文档的方式，它们应该是等效的，但显然情况并非如此。

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> …