如何去掉(移出)包含大数据结构类型的结构字段?

顺便说一下,你已经写过的作品。这转让了 a.v 到 b 使无效 a (旧变量不再存在)。这意味着我们不需要用有效的数据替换取出的数据。这是拥有所有权相对于可变访问的好处之一。

let a = A {
    v: VeryLargeStructure {/* ... */},
};
let b = a.v;
// Now `b` has ownership of what was once `a.v`.
// `a` no longer exists as a single entity,
// but you could extract other fields too if you wish.
// E.g. `let c = a.w` if `w` is some other field of `a`.

在上下文中,

struct VeryLargeStructure {
    // many tens of fields here
}

struct A {
    v: VeryLargeStructure,
}

fn main() {
    let a = A {
        v: VeryLargeStructure {/* ... */},
    };
    let b = a.v;
}

(playground)

有点脱色的版本是 destructuring . 它可以方便地同时提取多个字段。不过,就一个领域而言, let b = a.v; 可能是首选,因为它更简单、更清晰。

struct VeryLargeStructure {
    // many tens of fields here
}

struct A {
    v: VeryLargeStructure,
}

fn main() {
    let a = A {
        v: VeryLargeStructure {/* ... */},
    };
    let A { v: b } = a;
}

(playground)

注意,如果 A 工具 Drop ,它必须保留所有油田的所有权。这就意味着你不能不更换就搬出去。替换数据(例如 mem::replace )不过,很好。