可变的借用不一致

发布于 2025-01-21 07:03:45 字数 662 浏览 1 评论 0原文

以下program_pass在生锈中编译。

fn main() {
    let mut x = 0;
    let mut y = &mut x;
    let mut z = &mut y;
    let mut last = &mut z;

    let mut alt_y = &mut x;
    let mut alt_z = &mut alt_y;

    z = &mut alt_y;  // *last = &mut alt_y;

}

以下program_error没有。

fn main() {
    let mut x = 0;
    let mut y = &mut x;
    let mut z = &mut y;
    let mut last = &mut z;

    let mut alt_y = &mut x;
    let mut alt_z = &mut alt_y;

    *last = &mut alt_y;   // z = &mut alt_y;

}

program_error中违反了什么，而不在program_pass中？刚刚开始，但这确实与我认为是我对锈蚀的理解有关。

原文

The following program_pass compiles in Rust.

fn main() {
    let mut x = 0;
    let mut y = &mut x;
    let mut z = &mut y;
    let mut last = &mut z;

    let mut alt_y = &mut x;
    let mut alt_z = &mut alt_y;

    z = &mut alt_y;  // *last = &mut alt_y;

}

The following program_error does not.

fn main() {
    let mut x = 0;
    let mut y = &mut x;
    let mut z = &mut y;
    let mut last = &mut z;

    let mut alt_y = &mut x;
    let mut alt_z = &mut alt_y;

    *last = &mut alt_y;   // z = &mut alt_y;

}

What is violated in program_error, while not in program_pass?
Just started, but this is really going against what I thought to be my understanding of Rust.

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笑饮青盏花 2025-01-28 07:03:45

不是不一致，它是预期的行为。
在第一种情况下，没有使用可变的参考。实际上没有问题，因此生锈的编译器很高兴。

在第二种情况下，Rust编译器认为可变参考最后被删除，因此将其作为价值访问。众所周知，Rust不允许两个可变的借款。

参考： deref

证明我的观点已经调整了您的计划位

fn main() {
    let mut x = 0;
    let mut y = &mut x;
    let mut z = &mut y;
    let mut last = &mut z;

    let mut alt_y = &mut x;
    let mut alt_z = &mut alt_y;

    // notice the RHS here assigning 
    // mutable reference of i32 literal
    *last = &mut &mut 4;
    //                ^ not related to x anyhow
}

和现在的错误将揭示问题背后的原因\

error[E0499]: cannot borrow `x` as mutable more than once at a time
  --> src/main.rs:7:21
   |
3  |     let mut y = &mut x;
   |                 ------ first mutable borrow occurs here
...
7  |     let mut alt_y = &mut x;
   |                     ^^^^^^ second mutable borrow occurs here
...
11 |     *last = &mut &mut 4;
   |     ------------------- first borrow later used here

It's not inconsistency, it is is a intended behavior instead.
In first case, no mutable reference is being used. Effectively no issue, so rust compiler is happy.

In second case the rust compiler sees that mutable reference last is being dereferenced so it is taken as value access. And as we know rust doesn't allow the two mutable borrows.

Reference: Deref

To prove my point have tweaked your program a bit

fn main() {
    let mut x = 0;
    let mut y = &mut x;
    let mut z = &mut y;
    let mut last = &mut z;

    let mut alt_y = &mut x;
    let mut alt_z = &mut alt_y;

    // notice the RHS here assigning 
    // mutable reference of i32 literal
    *last = &mut &mut 4;
    //                ^ not related to x anyhow
}

And now the error will reveal the reason behind the issue\

error[E0499]: cannot borrow `x` as mutable more than once at a time
  --> src/main.rs:7:21
   |
3  |     let mut y = &mut x;
   |                 ------ first mutable borrow occurs here
...
7  |     let mut alt_y = &mut x;
   |                     ^^^^^^ second mutable borrow occurs here
...
11 |     *last = &mut &mut 4;
   |     ------------------- first borrow later used here