分配给借入变量如何违反引用规则？ [英] How does assigning to a borrowed variable violate the rules of references?

问题描述

我有以下代码：

struct Foo<'a> {
    link: &'a i32,
}

fn main() {
    let mut x = 33;
    println!("x:{}", x);
    let ff = Foo { link: &x };
    x = 22;
}

哪个会生成此编译器错误：

Which generates this compiler error:

error[E0506]: cannot assign to `x` because it is borrowed
 --> src/main.rs:9:5
  |
8 |     let ff = Foo { link: &x };
  |                           - borrow of `x` occurs here
9 |     x = 22;
  |     ^^^^^^ assignment to borrowed `x` occurs here

Rust书只有两个规则：

The Rust book has only two rules:

1. 对资源的一个或多个引用（& T ），


2. 恰好是一个可变引用（& mut T ）。

1. one or more references (&T) to a resource,
2. exactly one mutable reference (&mut T).

我有一个可变变量和一个不可变链接。

I have one mutable variable and one immutable link. Why does the compiler give an error?

推荐答案

Rust编程语言 定义了引用规则：

• 在任何给定时间，您都可以具有一个可变引用或任意数量的不可变引用。


• 引用必须始终有效。

隐式重新分配变量需要可变的引用：

Reassigning a variable implicitly requires a mutable reference:

fn main() {
    let mut x = 33;
    let link = &x;
    x = 22;
    *(&mut x) = 22; // Basically the same thing
}

重要的是，重新分配了变量 mutates 变量，该变量将导致不可变引用链接的值发生更改，这是不允许的。

Importantly, reassigning a variable mutates the variable, which would cause the value of the immutable reference link to change, which is disallowed.

请注意，变量的初始赋值不要求变量是可变的：

Note that the initial assignment of the variable does not require the variable to be mutable:

fn main() {
    let x;
    // Some other code
    x = 42;
}

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