我正在尝试做一个简单的分配器,它从一个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并使cell1只对我的单元格池可用时,就会发生以下错误:
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是否有人对如何修复此代码有建议,以便它可以在空闲列表中编译并可能包含2+项?
我希望填充数组的引用列表,然后能够弹出它们--使用它们一段时间,并将已使用/完成的值返回给自由职业者。
这里的目的是构建一个使用#![nostd]指令的库,因此它需要一个分配器接口才能正常运行。
发布于 2016-03-27 06:26:28
多亏了starblue的指针,我决定强制分配程序和单元格的生命周期是相同的,方法是将它们放置到一个结构中,并将其放入堆栈中。最后的结果是:
// 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)
}
}我需要添加哨兵结构,以避免在执行多个分配时重复借用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将哨兵存储在Alloc中后,我可以保证在函数返回后永远不会修改glob。
发布于 2016-03-26 14:38:59
问题是,您总是使用相同的生存期'a。这迫使cell0和cell1具有相同的生存期,这是不可能的,因为必须先定义一个。如果仔细阅读错误消息,可以看到它抱怨第二个单元格的生存期,不包括定义第一个单元格的语句。
我不知道这是一个错误或错误,它严格地实现了生命周期,或者它是否是生命类型系统中固有的(我还没有看到一个正式的定义)。
我也不知道怎么解决它。通过引入额外的生存期变量,我在一些示例代码中修正了类似的问题,但我无法使它对您的代码有效。
https://stackoverflow.com/questions/36233735
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