如何在不复制的情况下将字符串值从数组移到元组？ [英] How do I move String values from an array to a tuple without copying?

问题描述

我有一个固定大小的数组，该数组的大小为 String s： [String; 2] 。我想将其转换为（String，String）。

I have a fixed size array of Strings: [String; 2]. I want to turn it into a (String, String). Can I do this without copying the values?

我正在特别处理的代码段如下：

The piece of code that I'm working on in particular is the following:

let (basis, names_0, names_1) = if let Some(names) = self.arg_name {
    (ComparisonBasis::Name, names[0], names[1])
} else {
    (ComparisonBasis::File, self.arg_file[0], self.arg_file[1])
};

类型：

self.arg_name: Option<[String; 2]>
self.arg_file: Vec<String>

现在我遇到了错误

cannot move out of type `[std::string::String; 2]`, a non-copy fixed-size array [E0508]

和

cannot move out of indexed content [E0507]

用于 if

推荐答案

您已经省略了很多上下文，所以我在几个方面进行了猜测。我也更想知道您所提出的问题，而不是您的摘录所隐含的问题。

You've omitted a fair amount of context, so I'm taking a guess at a few aspects. I'm also hewing a little closer to the question you asked, rather than the vaguer one implied by your snippets.

struct NeverSpecified {
    arg_names: Option<[String; 2]>,
    arg_file: Vec<String>,
}

impl NeverSpecified {
    fn some_method_i_guess(mut self) -> (String, String) {
        if let Some(mut names) = self.arg_names {
            use std::mem::replace;
            let name_0 = replace(&mut names[0], String::new());
            let name_1 = replace(&mut names[1], String::new());
            (name_0, name_1)
        } else {
            let mut names = self.arg_file.drain(0..2);
            let name_0 = names.next().expect("expected 2 names, got 0");
            let name_1 = names.next().expect("expected 2 names, got 1");
            (name_0, name_1)
        }
    }
}

我使用 std :: mem :: replace 切换数组的内容，同时使其处于有效状态。这是必要的，因为Rust不允许您使用部分有效的数组。此路径没有任何副本或分配。

I use std::mem::replace to switch the contents of the array, whilst leaving it in a valid state. This is necessary because Rust won't allow you to have a "partially valid" array. There are no copies or allocations involved in this path.

在另一路径中，我们必须手动将元素从向量中拉出。同样，您不能只是通过索引将值移出容器（实际上这是对整个索引的限制）。相反，我使用 Vec ::耗尽 基本上将向量中的前两个元素砍掉，然后从生成的迭代器中提取它们。要明确：此路径不涉及任何副本或分配，或。

In the other path, we have to pull elements out of the vector by hand. Again, you can't just move values out of a container via indexing (this is actually a limitation of indexing overall). Instead, I use Vec::drain to essentially chop the first two elements out of the vector, then extract them from the resulting iterator. To be clear: this path doesn't involve any copies or allocations, either.

顺便说一句，那些期望方法应该永远不会被触发（因为 drain 会进行边界检查），但是比起遗憾更好的偏执；如果您想用 unwrap（）调用代替它们，那应该没事。.

As an aside, those expect methods shouldn't ever be triggered (since drain does bounds checking), but better paranoid than sorry; if you want to replace them with unwrap() calls instead, that should be fine..

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