關於鏽蝕壽命的問題

Question

我試圖實現基於TypedArena的內存池。下面是我的原代碼的簡化版本：

#![feature(rustc_private)] 
extern crate arena; 
use arena::TypedArena; 

pub struct MemoryPool { 
    arena: TypedArena<Vec<u8>>, 
    bytes_allocated: usize, 
} 

impl MemoryPool { 
    pub fn consume(&mut self, buf: Vec<u8>) -> &[u8] { 
     self.bytes_allocated += buf.capacity(); 
     self.arena.alloc(buf) 
    } 
} 

pub struct ByteArray<'a> { 
    data: &'a [u8], 
} 

impl<'a> ByteArray<'a> { 
    pub fn set_data(&mut self, data: &'a [u8]) { 
     self.data = data; 
    } 
} 

pub struct S<'a> { 
    pool: &'a mut MemoryPool, 
} 

impl<'a> S<'a> { 
    pub fn write(&mut self, buffer: &mut ByteArray<'a>) { 
     let v = vec!(); 
     let data = self.pool.consume(v); 
     buffer.set_data(data); 
    } 
} 

然而，編譯器抱怨行：let data = self.pool.consume(v);：

error[E0495]: cannot infer an appropriate lifetime for autoref due to conflicting requirements 
    --> <anon>:34:26 
    | 
34 |  let data = self.pool.consume(v); 
    |       ^^^^^^^ 
    | 
note: first, the lifetime cannot outlive the anonymous lifetime #1 defined on the body at 32:54... 
    --> <anon>:32:55 
    | 
32 |  pub fn write(&mut self, buffer: &mut ByteArray<'a>) { 
    | _______________________________________________________^ starting here... 
33 | |  let v = vec!(); 
34 | |  let data = self.pool.consume(v); 
35 | |  buffer.set_data(data); 
36 | | } 
    | |___^ ...ending here 
note: ...so that reference does not outlive borrowed content 
    --> <anon>:34:16 
    | 
34 |  let data = self.pool.consume(v); 
    |    ^^^^^^^^^ 
note: but, the lifetime must be valid for the lifetime 'a as defined on the body at 32:54... 
    --> <anon>:32:55 
    | 
32 |  pub fn write(&mut self, buffer: &mut ByteArray<'a>) { 
    | _______________________________________________________^ starting here... 
33 | |  let v = vec!(); 
34 | |  let data = self.pool.consume(v); 
35 | |  buffer.set_data(data); 
36 | | } 
    | |___^ ...ending here 
note: ...so that types are compatible (expected &mut ByteArray<'_>, found &mut ByteArray<'a>) 
    --> <anon>:35:12 
    | 
35 |  buffer.set_data(data); 
    |   ^^^^^^^^ 

我的問題是：

1. 爲什麼data不有終身'a？我在想，由於pool的使用壽命爲a，並且consume的返回壽命與self相同，因此應該有終生使用期限'a。

2. 使代碼正常工作的最佳方式是什麼？基本上我想分配新的字節並將其生存期調整爲與內存池相同。我知道我可以直接使用TypedArena，因爲alloc不需要mut參考。不過，我真的想跟蹤其他信息，如bytes_allocated。

Answer 1

讓我們一步解決此步驟：

cannot infer an appropriate lifetime for autoref 

「autoref」描述爲建設一個方法的self說法正確的引用的過程。編譯器無法找到具有正確的生命週期的引用來調用consume()。爲什麼它不能？

note: first, the lifetime cannot outlive the anonymous lifetime #1 defined on the body at 32:54... 
    --> <anon>:32:55 
    | 
32 |  pub fn write(&mut self, buffer: &mut ByteArray<'a>) { 
    | _______________________________________________________^ starting here... 
33 | |  let v = vec!(); 
34 | |  let data = self.pool.consume(v); 
35 | |  buffer.set_data(data); 
36 | | } 
    | |___^ ...ending here 
note: ...so that reference does not outlive borrowed content 
    --> <anon>:34:16 
    | 
34 |  let data = self.pool.consume(v); 
    |    ^^^^^^^^^ 

「匿名生命週期＃1」是指生命週期爲&mut self。這個說明只是說：我們不能將壽命大於self的壽命的參考值傳遞給consume()：那麼consume()會認爲它的參數壽命比實際壽命長。

note: but, the lifetime must be valid for the lifetime 'a 

這是您已經預期應用的規則。但現在問題在哪裏？那麼：&mut self（匿名生命＃1）的壽命可以比生命短比'a！就這樣！我們可以很容易地解決這個問題：

impl<'a> S<'a> { 
    pub fn write<'b: 'a>(&'b mut self, buffer: &mut ByteArray<'a>) { 
     //  ^^^^^^^^ ^^ 
     ... 
    } 
} 

這裏我們僅舉先前匿名終身＃1能夠約束它，稱它有活得比'a（比'a壽命更長）。