如何通过获取可变变量的值来替换它的价值？ [英] How do you replace the value of a mutable variable by taking ownership of it?

问题描述

我正在使用 LinkedList ，我想删除所有未通过测试的元素。但是，我遇到错误无法移出借来的内容。

I am working with a LinkedList and I want to remove all elements which do not pass a test. However, I am running into the error cannot move out of borrowed content.

据我所知，这是因为我正在使用& mut self ，所以我没有权利使一个包含的值无效（即移动），甚至片刻构建一个新的列表它的值。

From what I understand, this is because I am working with &mut self, so I do not have the right to invalidate (i.e. move) one of the contained values even for a moment to construct a new list of its values.

在C ++ / Java中，我只是迭代列表并删除任何符合条件的元素。由于我没有找到删除，我将其解释为迭代，过滤和收集。

In C++/Java, I would simply iterate the list and remove any elements which match a criteria. As there is no remove that I have yet found, I have interpreted it as an iterate, filter, and collect.

目标是避免创建临时列表，克隆值，并需要 self 并返回一个新对象。我构建了一个产生相同错误的例子。 游乐场。

The goal is to avoid creating a temporary list, cloning values, and needing take self and return a "new" object. I have constructed an example which produces the same error. Playground.

use std::collections::LinkedList;

#[derive(Debug)]
struct Example {
    list: LinkedList<i8>,
    // Other stuff here
}

impl Example {
    pub fn default() -> Example {
        let mut list = LinkedList::new();
        list.push_back(-5);
        list.push_back(3);
        list.push_back(-1);
        list.push_back(6);
        Example { list }
    }

    // Simmilar idea, but with creating a new list
    pub fn get_positive(&self) -> LinkedList<i8> {
        self.list.iter()
            .filter(|&&x| x > 0)
            .map(|x| x.clone())
            .collect()
    }

    // Now, attempt to filter the elements without cloning anything
    pub fn remove_negative(&mut self) {
        self.list = self.list.into_iter()
            .filter(|&x| x > 0)
            .collect()
    }
}

fn main() {
    let mut e = Example::default();
    println!("{:?}", e.get_positive());
    println!("{:?}", e);
}

在我的实际情况中，我不能简单地使用包装对象，因为它需要从不同的地方引用并包含其他重要的值。

In my actual case, I cannot simply consume the wrapping object because it needs to be referenced from different places and contains other important values.

在我的研究中，我发现了一些不安全代码，这让我质疑是否可以构建一个安全函数来执行此操作，其方式与 std :: mem :: replace 。

In my research, I found some unsafe code which leads me to question if a safe function could be constructed to perform this action in a similar way to std::mem::replace.

推荐答案

您可以 std :: mem :: swap 带有临时值的字段，然后将其替换为您修改后的列表像这样。最大的缺点是创建了新的LinkedList。我不知道它有多贵。

You can std::mem::swap your field with a temp, and then replace it with your modified list like this. The big downside is the creation of the new LinkedList. I don't know how expensive that is.

pub fn remove_negative(&mut self) {
    let mut temp = LinkedList::new();
    std::mem::swap(&mut temp, &mut self.list);

    self.list = temp.into_iter()
         .filter(|&x| x > 0)
         .collect();
}

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