移动结构后，我对结构成员的引用如何仍然有效?

问题描述

我正在使用函数(名为 get)创建对结构成员的引用，然后使用另一个函数(名为 pr)移动结构，然后我取消引用之前创建的指针.

I'm creating a reference to a structure member using a function (named get), then I move the struct using another function (named pr), then I dereference the previously created pointer.

我在这里做错了吗(也就是意外工作)，或者我的参考依据某些规则仍然有效?

Am I in the wrong here (a.k.a. working by accident), or is my reference is still valid by some rule?

struct MyStruct {
    inner: i32,
}

fn get(a: &MyStruct) -> &i32 {
    return &a.inner;
}

fn pr(a: MyStruct) {
    println!("MyStruct {}", a.inner);
}

fn main() {
    println!("Hello, world!");
    let x = MyStruct { inner: 3 };
    let &i = get(&x);
    pr(x);
    println!("i {}", i);
}

Rust 游乐场 输出:

Hello, world!
MyStruct 3
i 3

推荐答案

let 表达式得到模式匹配和

The let expression gets pattern-matched and

let &i = get(&x); // i.e. a &i32

导致 i 被分配给 i32 并且由于 i32 是可复制的，所以不存在所有权冲突.

Results in i being assigned to i32 and since i32 is copyable, there is no ownership violation.

Rust 参考声明一个 let 语句引入了一组由模式给出的新变量"(source) 和模式由文字、解构数组或枚举构造函数、结构和元组、变量绑定规范的某种组合组成"(来源).

The Rust reference states that "a let statement introduces a new set of variables, given by a pattern" (source) and "patterns consist of some combination of literals, destructured arrays or enum constructors, structs and tuples, variable binding specifications" (source).

绑定的左侧，&i 不仅仅是一个文字，它告诉编译器它应该尝试对右侧表达式进行模式匹配.在这种情况下，它导致 i 指向可复制值 (i32) 而不是引用 (&i32).换句话说:

The left-hand side of the binding, &i is not just a literal, which tells the compiler that it should try to pattern-match against the right-hand side expression. In this case it results in i pointing to a copyable value (i32) and not a reference (&i32). In other words:

let &i = get(&x);

相当于

let i = *get(&x);

所以 x 没有被借用，pr(x) 仍然适用.

So x is not borrowed and pr(x) is still applicable.

这篇关于移动结构后，我对结构成员的引用如何仍然有效?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！