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How to borrow an unwrapped Option<T>? [duplicate]

I want to iterate over a vector using .iter_mut() and .map():

fn calculate_distances(planes : &mut Vec<Aeroplane>, x: f64, y: f64) {
    fn calculate_distance(x1: &f64, y1: &f64, x2: &f64, y2: &f64) -> f6 { ... }
    planes.iter_mut().map(|a| if a.position.is_some() {
        let pos: &Position = &a.position.unwrap();
        a.distance = Some(calculate_distance(&x, &y, &pos.latitude, &pos.longitude));
    });
}

Aeroplane contains instance of my Position struct:

struct Position {
    latitude: f64,
    longitude: f64,
}

In my understanding I'm just borrowing the position information not moving out anything, but the borrow-checker refuses my code:

error[E0507]: cannot move out of borrowed content
   --> src/main.rs:145:31
    |
  4 |         let pos: &Position = &a.position.unwrap();
    |                               ^ cannot move out of borrowed content

Where is my mistake?

like image 278
PEAR Avatar asked Jan 21 '26 14:01

PEAR


1 Answers

You are looking for Option::as_ref:

Converts from Option<T> to Option<&T>.

The following code solves your problem:

let pos = a.position.as_ref().unwrap();

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

Your code does not work, because as stated by turbulencetoo you try to move the data out of the Option and borrow the moved data.

However, in this case the better solution would be if let:

if let Some(ref pos) = a.position {
    a.distance = Some(calculate_distance(&x, &y, &pos.latitude, &pos.longitude));
}

See also:

  • How do I borrow a reference to what is inside an Option<T>?
  • How to unwrap a &Result<_,_>?
like image 52
Tim Diekmann Avatar answered Jan 23 '26 06:01

Tim Diekmann



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