Option Destructure - Rust Programming

An if let pattern can unpack a successful parse and keep the fallback path explicit.

Program

Play the program to choose raw text and see whether it becomes structured data.

option_destructure.rs

fn main() {
    let raw = ;
    let parsed = parse_pair(raw);
    if let Some((name, count)) = parsed {
        println!("{name}={count}");
    } else {
        println!("invalid");
    }
}

fn parse_pair(text: &str) -> Option<(&str, i32)> {
    let (name, value) = text.split_once(':')?;
    let count = value.parse().ok()?;
    Some((name, count))
}

fn main() {
    let raw = ;
    let parsed = parse_pair(raw);
    if let Some((name, count)) = parsed {
        println!("{name}={count}");
    } else {
        println!("invalid");
    }
}

fn parse_pair(text: &str) -> Option<(&str, i32)> {
    let (name, value) = text.split_once(':')?;
    let count = value.parse().ok()?;
    Some((name, count))
}

fn main() {
    let raw = ;
    let parsed = parse_pair(raw);
    if let Some((name, count)) = parsed {
        println!("{name}={count}");
    } else {
        println!("invalid");
    }
}

fn parse_pair(text: &str) -> Option<(&str, i32)> {
    let (name, value) = text.split_once(':')?;
    let count = value.parse().ok()?;
    Some((name, count))
}

if let `if let Some((name, count))` handles the success shape directly.

question mark `?` returns `None` early when splitting or parsing fails.

tuple pattern `(name, count)` destructures both fields from the successful result.