How to destructure and derefence tuple in rust

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Is there a way to simplify the below code by dereferencing foo and bar on the same line as we destructure the tuple?

let a = "a";
let b = "b";
let c = (&a, &b);

// Can we dereference here?
let (foo, bar) = c;

let foo_loc = *foo;
let bar_loc = *bar;
println!("{}", foo_loc);
println!("{}", bar_loc);
rust destructuring reference
1 Answers

You can pattern match the references:

fn main() {
    let a = "a";
    let b = "b";
    let c = (&a, &b);
    let (&foo, &bar) = c;
    println!("{}", foo);
    println!("{}", bar);
}

Playground

user_16951720
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