Can I destructure a tuple without binding the result to a new variable in a let/match/for statement?

Question

I\'d like to destructure a tuple and assign part of the result to a new variable and assign another part of the result to an existing.

The following code illustrates

Answer 1

No.

Destructuring is something you can only do with patterns; the left-hand side of an assignment is not a pattern, hence you can't destructure-and-assign.

See proto-RFC 372 (Destructuring assignment) which discusses the possibility of adding this feature.

Answer 2

Nightly Rust has feature(destructuring_assignment), which allows your original attempt to compile as-is:

#![feature(destructuring_assignment)]

fn main() {
    let mut list = &[0, 1, 2, 3][..];
    while !list.is_empty() {
        let head;
        (head, list) = list.split_at(1);
        println!("{:?}", head);
    }
}

[0]
[1]
[2]
[3]

---

However, I'd solve this using stable features like slice pattern matching, which avoids the need for the double check in split_at and is_empty:

fn main() {
    let mut list = &[0, 1, 2, 3][..];

    while let [head, rest @ ..] = list {
        println!("{:?}", head);
        list = rest;
    }
}

See also:

• How to swap two variables?