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

Question

Suppose I have a value x that has the type of Option<T>, how do I convert it to Option<&T> ?
I tried using map:

let result = x.map(|x| &x)
                       ^^ Error

But I got the error:

cannot return reference to function parameter `x`
returns a reference to data owned by the current function

Related: [How to borrow an unwrapped Option?](https://stackoverflow.com/questions/51385381/how-to-borrow-an-unwrapped-optiont) — Sven Marnach, Feb 01 '21 at 10:26
Does this answer your question? [Cannot move out of value which is behind a shared reference when unwrapping](https://stackoverflow.com/questions/32338659/cannot-move-out-of-value-which-is-behind-a-shared-reference-when-unwrapping) — pretzelhammer, Feb 01 '21 at 12:54
@pretzelhammer No it does not, although the solution looks similar, I believe it's a totally different question. I'm specifically asking how to convert Option to Option<&T>, even the error thrown is phrased differently. Therefore I request that this question to be opened again. — Wong Jia Hau, Feb 01 '21 at 13:06

score 5 · Accepted Answer · answered Feb 01 '21 at 03:24

5

After some trial and error, I found the method, which is as_ref.

answered Feb 01 '21 at 03:24

Wong Jia Hau

1 Answers1