我可以将可变切片引用重新分配给自身的子切片吗？

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 不返回值并且只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];
    }
}

给了

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];
   |                     ^^^^^^^^^^^^^^^^^^^^^^^^

有没有办法让这个工作，或者我需要更明确地跟踪我感兴趣的切片的边界？

Answer 1

我稍微修改了 Masklinn 的代码，以允许在同一个堆栈上调用多个 .pop()s：

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（为了明确起见，我添加了类型注释）：

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

我们暂时将self.data 替换为一个空切片，以便我们可以拆分切片。如果你写得简单

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，问题在于self.data 是可变引用，而可变引用不是Copy（记住，一次只能有一个）。而且您不能离开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，它用&mut self的生命周期“感染”它。这是一条死胡同：我们希望引用 live for 'a，但它实际上只在 &mut self 的生命周期内有效，而这些生命周期通常是不相关的（并且它们不需要相关），这让编译器感到困惑。

为了帮助编译器，我们使用std::mem::replace 将切片显式移出self.data，并暂时将其替换为空切片which can be any lifetime。现在我们可以用data 做任何事情，而不用纠结&mut self 的生命周期。

Answer 2

对于子问题 2，您需要指出 &mut self 和 'a 之间的关系，否则它们被视为不相关。我不知道是否存在通过生命周期省略的捷径，但如果您指定 self 为 'a 生存，那么您就可以了。

对于子问题 1，编译器不会“看穿”函数调用（包括对函数调用去糖的索引），因此它不知道 &self.data[n - 1] 和 &mut self.data[0..n-1] 不重叠。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