如何解开 &Result<_,_>? [英] How to unwrap a &Result<_,_>?

问题描述

从 &Result 类型中提取数据的好方法是什么?

What is a good way to extract data from a &Result type?

在我的特定情况下，我有一个 &Result 类型，我无法打开它，因为我不拥有该对象.我试图取消引用并克隆它(*left_item).clone()，但这只是给我一个错误的注释:

In my specific case, I have a &Result<DirEntry, Error> type, which I can't unwrap because I don't own the object. I tried to dereference and clone it (*left_item).clone(), but that just gives me a error with the note:

the method `clone` exists but the following trait bounds were not satisfied:
`std::result::Result<std::fs::DirEntry, std::io::Error> : std::clone::Clone`

推荐答案

您正在寻找 Result::as_ref:

You are looking for Result::as_ref:

从 Result 转换为 Result<&T, &E>.

产生一个新的Result，其中包含对原始数据的引用，保留原始数据.

Produces a new Result, containing a reference into the original, leaving the original in place.

以下代码可以解决您的问题:

The following code solves your problem:

let entry: &DirEntry = result.as_ref().unwrap();

对于可变版本，结果::as_mut 已提供.

For a mutable version, Result::as_mut is provided.

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