How to convert Option<&T> to Option<T> in the most idiomatic way in Rust?

Question

When using HashMap's get method, I get an Option<&T>, I've encountered it again this time with Option<&String>. I'd like to get an owned value Option<String>. Is this possible without me writing map(|x| x.to_owned())?

I'm just wondering if there's a way to write a blanket implementation for any of the utility traits to achieve that?

Answer 1

Option comes with utility methods for various transformations, which are listed in its documentation. For any T that implements Clone (which String does), Option<&T>::cloned does what you're looking for.

Clone is more specific than ToOwned, so .cloned() isn't an exact match for .map(|x| x.to_owned()). For example, it won't turn an Option<&str> into an Option<String>; for that you will have to stick with map.

Since Rust 1.35, when T is Copy, .copied() does the same thing as .cloned(), but it will fail to compile when T is not Copy. You might use this when you want to be explicit that the clone is cheap.

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See also:

• How to clone last element from vector?
• Get the last element of a vector and push it to the same vector