rust - 为什么可以从函数返回对文字的可变引用？

Question

当前版本The Rustonomicon有这个示例代码:

use std::mem;

pub struct IterMut<'a, T: 'a>(&'a mut [T]);

impl<'a, T> Iterator for IterMut<'a, T> {
    type Item = &'a mut T;

    fn next(&mut self) -> Option<Self::Item> {
        let slice = mem::replace(&mut self.0, &mut []);
        if slice.is_empty() {
            return None;
        }

        let (l, r) = slice.split_at_mut(1);
        self.0 = r;
        l.get_mut(0)
    }
}

我特别对这条线感到困惑:

let slice = mem::replace(&mut self.0, &mut []);
//                                    ^^^^^^^ 

这个借书怎么查？如果这是一个不可变的借用， RFC 1414表示 []右值应该有 'static生命周期，因此不可变借用将借用检查，但该示例显示了可变借用!似乎必须发生两件事之一:

要么[]是临时的(以便它可以可变地使用)，在这种情况下它不会有 'static终生不得借阅；

或者那个[]有 'static生命周期，因此不应该进行可变借用(因为我们在借用时不保证独占访问)，并且不应该借用检查。

我错过了什么？

有关的:

Why can I return a reference to a local literal but not a variable?

这个问题侧重于不可变引用；这个问题是关于可变 引用。

Why is it legal to borrow a temporary?

这个问题的重点是在函数内部获取引用；这个问题是关于返回 一个引用。

Answer 1

TL;DR:空数组在编译器中是特殊情况，它是安全的，因为你永远不能解引用零长度数组的指针，所以不可能有可变别名。

RFC 1414 ，右值静态提升，讨论将值提升到 static 的机制值。它有一个关于 mutable references 可能的扩展的部分(粗体我的):

It would be possible to extend support to &'static mut references, as long as there is the additional constraint that the referenced type is zero sized.

This again has precedence in the array reference constructor:

// valid code today
let y: &'static mut [u8] = &mut [];

The rules would be similar:

• If a mutable reference to a constexpr rvalue is taken. (&mut <constexpr>)
• And the constexpr does not contain a UnsafeCell { ... } constructor.
• And the constexpr does not contain a const fn call returning a type containing a UnsafeCell.
• And the type of the rvalue is zero-sized.
• Then instead of translating the value into a stack slot, translate it into a static memory location and give the resulting reference a 'static lifetime.

The zero-sized restriction is there because aliasing mutable references are only safe for zero sized types (since you never dereference the pointer for them).

由此，我们可以看出对空数组的可变引用目前在编译器中是特殊情况。在 Rust 1.39 中，所讨论的扩展尚未实现:

struct Zero;

fn example() -> &'static mut Zero {
    &mut Zero
}

error[E0515]: cannot return reference to temporary value
 --> src/lib.rs:4:5
  |
4 |     &mut Zero
  |     ^^^^^----
  |     |    |
  |     |    temporary value created here
  |     returns a reference to data owned by the current function

虽然阵列版本确实有效:

fn example() -> &'static mut [i32] {
    &mut []
}

也可以看看:

Why is it legal to borrow a temporary?

Why can I return a reference to a local literal but not a variable?