CODEWITHSUNDEEP
rustmutable
xixixao
xixixao

Reputation: 563

How can I override all the fields in a mutable reference using another struct?

How can I avoid listing all the fields when using x to populate input?

struct StructX {
    a: u32,
    b: u32,
}

trait TraitY {
    fn foo(info: &mut StructX) -> bool;
}

impl TraitY for SomeZ {
    fn foo(input: &mut StructX) -> bool {
        let mut x = StructX { /*....*/ };
        // do something with x, then finally:
        input.a = x.a;
        input.b = x.b;
    }
}

In C++ it would be just input = x, but that doesn't work in Rust. Note that this is an "interface" so I cannot change the type of input to something else.

Upvotes: 1

Views: 457

Answers (1)

MaxV
MaxV

Reputation: 2800

You have to dereference input (playground):

struct StructX {
    a: u32,
    b: u32,
}

trait TraitY {
    fn foo(info: &mut StructX) -> bool;
}

impl TraitY for SomeZ {
    fn foo(input: &mut StructX) -> bool {
        let mut x = StructX { /*....*/ };
        // do something with x, then finally:
        *input = x;

        return true;
    }
}

If you wouldn't like to move x into input then you can use Clone::clone_from

playground

#[derive(Clone)]
struct StructX {
    a: u32,
    b: u32,
}


trait TraitY {
    fn foo(info: &mut StructX) -> bool;
}

struct SomeZ{}

impl TraitY for SomeZ {
    fn foo(input: &mut StructX) -> bool {
        let mut x = StructX { a:42, b:56};
        x.a = 43;
        input.clone_from(&x);
        
        return true;
    }
}

Upvotes: 6

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