我可以将可变切片引用重新分配给它自己的子切片吗？

Question

我正在实现一个类似堆栈的结构，其中该结构包含对切片的可变引用。

struct StackLike<'a, X> {
    data: &'a mut [X],
}

我希望能够从堆栈中弹出最后一个元素，例如：

impl<'a, X> StackLike<'a, X> {
    pub fn pop(&mut self) -> Option<&'a X> {
        if self.data.is_empty() {
            return None;
        }
        let n = self.data.len();
        let result = &self.data[n - 1];
        self.data = &mut self.data[0..n - 1];
        Some(result)
    }
}

这失败了：

error[E0495]: cannot infer an appropriate lifetime for lifetime parameter in function call due to conflicting requirements
  --> src/lib.rs:11:23
   |
11 |         let result = &self.data[n - 1];
   |                       ^^^^^^^^^^^^^^^^
   |
note: first, the lifetime cannot outlive the anonymous lifetime #1 defined on the method body at 6:5...
  --> src/lib.rs:6:5
   |
6  | /     pub fn pop(&mut self) -> Option<&'a X> {
7  | |         if self.data.is_empty() {
8  | |             return None;
9  | |         }
...  |
13 | |         Some(result)
14 | |     }
   | |_____^
note: ...so that reference does not outlive borrowed content
  --> src/lib.rs:11:23
   |
11 |         let result = &self.data[n - 1];
   |                       ^^^^^^^^^
note: but, the lifetime must be valid for the lifetime `'a` as defined on the impl at 5:6...
  --> src/lib.rs:5:6
   |
5  | impl<'a, X> StackLike<'a, X> {
   |      ^^
note: ...so that the expression is assignable
  --> src/lib.rs:13:9
   |
13 |         Some(result)
   |         ^^^^^^^^^^^^
   = note: expected  `std::option::Option<&'a X>`
              found  `std::option::Option<&X>`

即使 a simplified version of pop 没有 return 值且仅缩小切片也不起作用。

impl<'a, X> StackLike<'a, X> {
    pub fn pop_no_return(&mut self) {
        if self.data.is_empty() {
            return;
        }
        let n = self.data.len();
        self.data = &mut self.data[0..n - 1];
    }
}

这给出了

error[E0495]: cannot infer an appropriate lifetime for lifetime parameter in function call due to conflicting requirements
  --> src/lib.rs:11:26
   |
11 |         self.data = &mut self.data[0..n - 1];
   |                          ^^^^^^^^^^^^^^^^^^^
   |
note: first, the lifetime cannot outlive the anonymous lifetime #1 defined on the method body at 6:5...
  --> src/lib.rs:6:5
   |
6  | /     pub fn pop_no_return(&mut self) {
7  | |         if self.data.is_empty() {
8  | |             return;
9  | |         }
10 | |         let n = self.data.len();
11 | |         self.data = &mut self.data[0..n - 1];
12 | |     }
   | |_____^
note: ...so that reference does not outlive borrowed content
  --> src/lib.rs:11:26
   |
11 |         self.data = &mut self.data[0..n - 1];
   |                          ^^^^^^^^^
note: but, the lifetime must be valid for the lifetime `'a` as defined on the impl at 5:6...
  --> src/lib.rs:5:6
   |
5  | impl<'a, X> StackLike<'a, X> {
   |      ^^
note: ...so that reference does not outlive borrowed content
  --> src/lib.rs:11:21
   |
11 |         self.data = &mut self.data[0..n - 1];
   |                     ^^^^^^^^^^^^^^^^^^^^^^^^

有没有办法使这项工作起作用，或者我是否需要更明确地跟踪我感兴趣的切片的边界？

Answer 1

子题2需要说明&mut self和'a之间的关系，否则视为不相关。我不知道是否有通过生命周期省略的捷径，但如果你指定 self 活到 'a 你就没问题。

对于子问题 1，编译器不会 "see through" 函数调用（包括索引 which desugars to a function call），所以它不知道 &self.data[n - 1] 和 &mut self.data[0..n-1] 是不重叠的。您需要使用 split_mut_last.

struct StackLike<'a, X> {
    data: &'a mut [X],
}

impl<'a, X> StackLike<'a, X> {
    pub fn pop(&'a mut self) -> Option<&'a X> {
        if let Some((last, subslice)) = self.data.split_last_mut() {
            self.data = subslice;
            Some(last)
        } else {
            None
        }
    }
}

playground

Answer 2

我稍微修改了 Masklinn 的代码以允许在同一堆栈上调用多个 .pop()：

struct StackLike<'a, X> {
    data: &'a mut [X],
}

impl<'a, X> StackLike<'a, X> {
    pub fn pop(&mut self) -> Option<&'a mut X> {
        let data = std::mem::replace(&mut self.data, &mut []);
        if let Some((last, subslice)) = data.split_last_mut() {
            self.data = subslice;
            Some(last)
        } else {
            None
        }
    }
}

fn main() {
    let mut data = [1, 2, 3, 4, 5];
    let mut stack = StackLike { data: &mut data };

    let x = stack.pop().unwrap();
    let y = stack.pop().unwrap();
    println!("X: {}, Y: {}", x, y);
}

这里棘手的部分是这一行（我添加了一个类型注释以明确性）：

let data: &'a mut [X] = std::mem::replace(&mut self.data, &mut []);

我们暂时用一个空切片替换self.data，这样我们就可以分割切片了。如果你简单地写

let data: &'a mut [X] = self.data;

编译器会不高兴：

error[E0312]: lifetime of reference outlives lifetime of borrowed content...
  --> src/main.rs:7:33
   |
7  |         let data: &'a mut [X] = self.data;
   |                                 ^^^^^^^^^
   |
note: ...the reference is valid for the lifetime `'a` as defined on the impl at 5:6...
  --> src/main.rs:5:6
   |
5  | impl<'a,  X> StackLike<'a, X> {
   |      ^^
note: ...but the borrowed content is only valid for the anonymous lifetime #1 defined on the method body at 6:5
  --> src/main.rs:6:5
   |
6  | /     pub fn pop(&mut self) -> Option<&'a mut X> {
7  | |         let data: &'a mut [X] = self.data;
8  | |         if let Some((last, subslice)) = data.split_last_mut() {
9  | |             self.data = subslice;
...  |
13 | |         }
14 | |     }
   | |_____^

据我了解，问题是 self.data 是一个可变引用，而可变引用不是 Copy （记住，你一次只能有一个）。而且您不能移出 self.data，因为 self 是可变引用，而不是所有者。所以编译器试图做的是重新借用 self.data，"infects" 它的生命周期为 &mut self。这是一个死胡同：我们希望引用存活 'a，但它实际上只在 &mut self 的生命周期内有效，而且这些生命周期通常是不相关的（而且它们不需要相关），这让编译器感到困惑。

为了帮助编译器，我们使用 std::mem::replace 显式地将切片移出 self.data 并临时用一个空切片替换它。现在我们可以用 data 做任何事情，而不会与 &mut self.

的生命周期纠缠在一起

我可以将可变切片引用重新分配给它自己的子切片吗？

Can I reassign a mutable slice reference to a sub-slice of itself?

lifetime

rust

borrowing