How do I destructure a tuple so that the bindings are mutable?

问题

If I have the following struct:

struct MyStruct { tuple: (i32, i32) };

And the following function:

// This will not compile
fn function(&mut struct: MyStruct) {
    let (val1, val2) = struct.tuple;
    val1 = 1;
    val2 = 2;
}

How do I borrow val1 and val2 as mutable so when I reassign them the changes appear in the original struct?

回答1:

You've got a few problems:

You've put the &mut in the wrong place; &mut is part of the type, not the argument (unless you're destructuring the argument, which you aren't).
You can't call the argument struct, because that's a keyword.
You can't assign to a mutable reference with straight assignment.

So, with those in mind, here's a working solution:

#[derive(Debug)]
struct MyStruct {
    tuple: (i32, i32),
}

fn function(s: &mut MyStruct) {
    let (ref mut val1, ref mut val2) = s.tuple;
    *val1 = 1;
    *val2 = 2;
}

fn main() {
    let mut s = MyStruct { tuple: (0, 0) };
    function(&mut s);
    println!("{:?}", s);
}

The key here is that ref in a pattern binds by-reference; combining that with mut gives you a mutable reference. Specifically, it gives you a pair of &mut i32s. Since these are references, you have to de-reference them in order to assign through them (otherwise, you'd be trying to re-assign the reference itself).

回答2:

You have two slightly different questions.

You can create a mutable bind by saying mut twice:

fn main() {
    let a = (1, 2);
    let (mut b, mut c) = a;
    b += 1;
    c += 2;

    println!("{}, {}", b, c);
}

But to have it change in the original tuple, you need a mutable reference into that tuple:

fn main() {
    let mut a = (1, 2);

    {
        let (ref mut b, ref mut c) = a;
        *b += 1;
        *c += 2;
        // Let mutable borrows end
    }

    println!("{:?}", a);
}

来源：https://stackoverflow.com/questions/31595087/how-do-i-destructure-a-tuple-so-that-the-bindings-are-mutable

标签

rust