简单资源分配器结构的生命周期错误

Question

我正在尝试制作一个简单的分配器，用于从固定的缓冲区池中分配和释放缓冲区。

struct AllocatedMemory<'a> {
    mem: &'a mut [u8],
    next: Option<&'a mut AllocatedMemory<'a>>,
}

struct Alloc<'a> {
    glob: Option<&'a mut AllocatedMemory<'a>>,
}

impl<'a> Alloc<'a> {
    fn alloc_cell(mut self: &mut Alloc<'a>) -> &mut AllocatedMemory<'a> {
        let rest: Option<&'a mut AllocatedMemory<'a>>;
        match self.glob {
            Some(ref mut root_cell) => {
                rest = std::mem::replace(&mut root_cell.next, None);
            }
            None => rest = None,
        }
        match std::mem::replace(&mut self.glob, rest) {
            Some(mut root_cell) => {
                return root_cell;
            }
            None => panic!("OOM"),
        }
    }

    fn free_cell(mut self: &mut Alloc<'a>, mut val: &'a mut AllocatedMemory<'a>) {
        match std::mem::replace(&mut self.glob, None) {
            Some(mut x) => {
                let discard = std::mem::replace(&mut val.next, Some(x));
                let rest: Option<&'a mut AllocatedMemory<'a>>;
            }
            None => {}
        }
        self.glob = Some(val);
    }
}

fn main() {
    let mut buffer0: [u8; 1024] = [0; 1024];
    let mut buffer1: [u8; 1024] = [0; 1024];
    {
        let mut cell1: AllocatedMemory = AllocatedMemory {
            mem: &mut buffer1[0..1024],
            next: None,
        };
        let mut cell0: AllocatedMemory = AllocatedMemory {
            mem: &mut buffer0[0..1024],
            next: None,
        };
        let mut allocator = Alloc { glob: None };
        allocator.free_cell(&mut cell1); //populate allocator with a cell
        allocator.free_cell(&mut cell0); //populate allocator with another cell (why does this fail?)

        let mut x = allocator.alloc_cell();
        allocator.free_cell(x);
        let mut y = allocator.alloc_cell();
        let mut z = allocator.alloc_cell();
        allocator.free_cell(y);
        allocator.free_cell(z);
    }
}

错误是

error: `cell0` does not live long enough
     allocator.free_cell(&mut cell0); //populate allocator with another cell (why does this fail?)

当我简单地删除 cell0 并且只有 cell1 可用于我的单元池时，会发生以下错误：

error: allocator does not live long enough
         let mut x = allocator.alloc_cell();
                     ^~~~~~~~~
note: reference must be valid for the block suffix following statement 0 at 46:69...
                                                          next: None};
         let mut cell0 : AllocatedMemory = AllocatedMemory{mem: &mut buffer0[0..1024],
                                                          next: None};
         let mut allocator = Alloc {glob : None};
         allocator.free_cell(&mut cell1); //populate allocator with a cell
         //allocator.free_cell(&mut cell0); //populate allocator with another cell (why does this fail?)

note: ...but borrowed value is only valid for the block suffix following statement 2 at 49:48
         let mut allocator = Alloc {glob : None};
         allocator.free_cell(&mut cell1); //populate allocator with a cell
         //allocator.free_cell(&mut cell0); //populate allocator with another cell (why does this fail?)

         let mut x = allocator.alloc_cell();
         allocator.free_cell(x);
               ...
error: aborting due to previous error

有没有人建议如何修复此代码，以便它编译并在空闲列表中可能有 2 个以上的项目？

我想填充对数组的引用列表，然后能够弹出它们 - 使用它们一段时间并将 used/finished 值放回空闲列表。

这里的动机是构建一个使用 #![nostd] 指令的库，因此它需要一个分配器接口才能正常运行。

Answer 1

问题是您总是使用相同的生命周期 'a。这迫使 cell0 和 cell1 具有相同的生命周期，这是不可能的，因为必须先定义一个。如果您仔细阅读错误消息，您会发现它抱怨第二个单元格的生命周期不包括定义第一个单元格的语句。

我不知道这是否是严格执行生命周期的错误或功能不当，或者它是否是生命周期类型系统中固有的（我还没有看到正式的定义）。

我也不知道怎么解决。我通过引入额外的生命周期变量解决了一些示例代码中的类似问题，但我无法使其适用于您的代码。

Answer 2

感谢 starblue 的指点，我决定强制分配器和单元格的生命周期相同，方法是将它们放入结构中并将该结构放入堆栈中。最终结果在这里：

// This code is placed in the public domain
struct AllocatedMemory<'a> {
   mem : &'a mut [u8],
   next : Option<&'a mut AllocatedMemory <'a> >,
}
struct Alloc<'a> {
  glob : Option<&'a mut AllocatedMemory <'a> >,
}

impl<'a> Alloc <'a> {
  fn alloc_cell(self : &mut Alloc<'a>) -> &'a mut AllocatedMemory<'a> {
      match self.glob {
        Some(ref mut glob_next) => {
             let rest : Option<&'a mut AllocatedMemory <'a> >;
             match glob_next.next {
                 Some(ref mut root_cell) => {
                    rest = std::mem::replace(&mut root_cell.next, None);
                 },
                 None => rest = None,
             }
             match std::mem::replace(&mut glob_next.next, rest) {
                Some(mut root_cell) =>
                {
                 return root_cell;
                },
                None => panic!("OOM"),
             }
        },
        None => panic!("Allocator not initialized"),
     }
  }
  fn free_cell(self : &mut Alloc<'a>,
               mut val : & 'a mut AllocatedMemory<'a>) {
      match self.glob {
          Some(ref mut glob_next) => {
              match std::mem::replace(&mut glob_next.next ,None) {
                  Some(mut x) => {
                      let _discard = std::mem::replace(&mut val.next, Some(x));
                  },
                  None => {},
              }
              glob_next.next = Some(val);
           },
           None => panic!("Allocator not initialized"),
      }
  }
}
struct AllocatorGlobalState<'a>{
  cell1 : AllocatedMemory<'a>,
  cell0 : AllocatedMemory<'a>,
  sentinel : AllocatedMemory<'a>,
  allocator :Alloc<'a>,

}
fn main() {
  let mut buffer0 : [u8; 1024] = [0; 1024];
  let mut buffer1 : [u8; 1024] = [0; 1024];
  let mut sentinel_buffer : [u8; 1] = [0];
  let mut ags : AllocatorGlobalState = AllocatorGlobalState {
  cell1 : AllocatedMemory{mem: &mut buffer1[0..1024],
                          next: None},
  cell0 : AllocatedMemory{mem: &mut buffer0[0..1024],
                          next: None},
  sentinel : AllocatedMemory{mem: &mut sentinel_buffer[0..1], next: None},
  allocator : Alloc {glob : None},
  };
  ags.allocator.glob = Some(&mut ags.sentinel);
  ags.allocator.free_cell(&mut ags.cell1);
  ags.allocator.free_cell(&mut ags.cell0);
  {
  let mut x = ags.allocator.alloc_cell();
  x.mem[0] = 4;
  let mut y = ags.allocator.alloc_cell();
  y.mem[0] = 4;
  ags.allocator.free_cell(y);
  let mut z = ags.allocator.alloc_cell();
  z.mem[0] = 8;
  //y.mem[0] = 5; // <-- this is an error (use after free)
  }
}

我需要添加 sentinel 结构以避免在进行多次分配时双重借用 ags.allocator。

cell.rs:65:19: 65:32 help: run `rustc --explain E0499` to see a detailed explanation
cell.rs:62:19: 62:32 note: previous borrow of `ags.allocator` occurs here; the mutable borrow prevents subsequent moves, borrows, or modification of `ags.allocator` until the borrow ends

随着哨兵存储在 Alloc 中，我可以保证我永远不会在函数 returns 之后修改 glob。

简单资源分配器结构的生命周期错误

Lifetime errors with a simple resource Allocator struct

lifetime

object-lifetime

rust

lifetime-scoping