Questions about Rust lifetime - rust

Question

I'm trying to implement a memory pool based on TypedArena. Here's a simplified version of my original code:
#![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);
}
}
However, the compiler complains about the line: 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);
| ^^^^^^^^
My question is:
Why data does not have lifetime 'a? I'm thinking that since pool has lifetime a and consume returns the same lifetime as self, it should have lifetime 'a.
What's the best way to make this code work as intended? Basically I want to allocate new bytes and adjust their lifetime to be the same as the memory pool. I know I can use TypedArena directly since alloc does not take a mut reference. However I really want to track other information such as bytes_allocated.

Answer 1

Let's tackle this step by step:
cannot infer an appropriate lifetime for autoref
"autoref" describes the process of building the right reference for the self argument of a method. The compiler is unable to find a reference with the right lifetime to call consume(). Why is it unable?
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);
| ^^^^^^^^^
The "anonymous lifetime #1" refers to the lifetime of &mut self. This note is just saying: we can't pass a reference with a lifetime greater than the lifetime of self into consume(): then consume() would think that its self argument lives longer than it actually does.
note: but, the lifetime must be valid for the lifetime 'a
This is the rule you already expected to be applied. But where is the problem now? Well: the lifetime of &mut self (anonymous lifetime #1) could life shorter than 'a! That's all! And we can fix it quite easily:
impl<'a> S<'a> {
pub fn write<'b: 'a>(&'b mut self, buffer: &mut ByteArray<'a>) {
// ^^^^^^^^ ^^
...
}
}
Here we just name the previously anonymous lifetime #1 to be able to bound it, saying that it has to outlive 'a (live longer than 'a).