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

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`
Tim Diekmann
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1 Answers1

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You are looking for Result::as_ref:

Converts from Result<T, E> to Result<&T, &E>.

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

The following code solves your problem:

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

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

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Tim Diekmann
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