如何在 Rust 中解构和取消引用元组

Question

有没有一种方法可以通过在解构元组时在同一行取消引用 foo 和 bar 来简化下面的代码？

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);

Answer 1

您可以模式匹配参考资料：

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

Playground

如何在 Rust 中解构和取消引用元组

How to destructure and derefence tuple in rust

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destructuring