How can I set structure members from a tuple without using match?

Question

I want to initialize two fields from a tuple returned from a function:

#[derive(Clone, Copy, Debug)]
struct Point {
    x: f64,
    y: f64,
    z: f64,
}

fn f() -> (f64, f64) {
    (5., 6.)
}

fn main() {
    let mut p = Point {
        x: 1.,
        y: 2.,
        z: 3.,
    };
    println!("{:?}", p);
    match f() {
        (x, y) => {
            p.x = x;
            p.y = y;
        }
    }
    println!("{:?}", p);
}

The obvious code (p.x, p.y) = f() did not compile so I have to use match. Do you have any idea how to make the match assignment simpler to read?

I want to

Call f only once per assignment of the two fields
Have f on the right side, and p.x and p.y and the left side, not how it is in match now

You could do `let (x,y) = f(); p.x = x; p.y = y;` . – Chris Emerson Aug 15 '16 at 23:09 — Chris Emerson, Aug 15 '16 at 23:09

score 1 · Answer 1 · answered Aug 15 '16 at 23:04

1

How's this?

let (x, y) = f();
p.x = x;
p.y = y;

You can put similar patterns in lets that you can put in matches.

answered Aug 15 '16 at 23:04

paholg

1,910
17
19

I expect something without introducing new variables – user1244932 Aug 15 '16 at 23:19
@user1244932 then the answer is, unfortunately, "no". – Shepmaster Aug 16 '16 at 02:31
@user1244932 you could put it in a block – paholg Aug 16 '16 at 05:41
How about a new method or function, so that you'd write `p.setxy(f())`? – Chris Emerson Aug 16 '16 at 07:07

How can I set structure members from a tuple without using match?

1 Answers1