为什么创建对解除引用的可变引用的可变引用有效? [英] Why does creating a mutable reference to a dereferenced mutable reference work?

问题描述

我知道您不允许在 Rust 中同时创建两个对一个对象的可变引用.我不完全理解为什么以下代码有效:

I understand you're not allowed to create two mutable references to an object at once in Rust. I don't entirely understand why the following code works:

fn main() {
    let mut string = String::from("test");
    let mutable_reference: &mut String = &mut string;
    mutable_reference.push_str(" test");
    // as I understand it, this creates a new mutable reference (2nd?)
    test(&mut *mutable_reference);

    println!("{}", mutable_reference);
}

fn test(s: &mut String) {
    s.push_str(" test");
}

推荐答案

规则

在任何时间点都只能有一个对特定值的可用可变引用.

There shall only be one usable mutable reference to a particular value at any point in time.

这不是空间排除(同一部分可以有多个引用)，而是时间排除.

This is NOT a spatial exclusion (there CAN be multiple references to the same piece) but a temporal exclusion.

机制

为了强制执行此操作，&mut T 不是 Copy；因此调用:

In order to enforce this, &mut T is NOT Copy; therefore calling:

test(mutable_reference);

应该将引用移动到test.

实际上这样做会使其在以后无法使用并且不符合人体工程学，因此 Rust 编译器会插入一个自动重新借用，就像您自己做的那样:

Actually doing this would make it unusable later on and not be very ergonomic, so the Rust compiler inserts an automatic reborrowing, much like you did yourself:

test(&mut *mutable_reference);

如果您愿意，可以强制移动:

You can force the move if you wanted to:

test({ let x = mutable_reference; x });

效果

再借，本质上就是借:

• mutable_reference 只要未命名的临时可变引用存在(或从它借用的任何东西)，就会被借用，
• 未命名的临时可变引用移入test，
• 在表达式处，未命名的临时可变引用被销毁，因此mutable_reference的借用结束.

• mutable_reference is borrowed for as long as the unnamed temporary mutable reference exists (or anything that borrows from it),
• the unnamed temporary mutable reference is moved into test,
• at the of expression, the unnamed temporary mutable reference is destroyed, and therefore the borrow of mutable_reference ends.

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