我有一个特征,该特征具有反序列化关联类型的功能。但是,该关联类型必须具有调用者决定的生存期,因此我有一个单独的特征,我使用了绑定到更高的特征,以便可以在任何生命周期中反序列化它。
我需要使用一个返回此关联类型的闭包。
我有以下代码可以做到这一点:
#![allow(unreachable_code)]
use std::marker::PhantomData;
trait Endpoint: for<'a> EndpointBody<'a> {}
trait EndpointBody<'a> {
type Out: 'a;
fn serialize(body: &Self::Out) -> Vec<u8>;
fn deserialize(raw_body: &'a [u8]) -> Self::Out;
}
// /////////////////////////////////////////////////////////
/// Trait object compatible handler
trait Handler {
fn execute(&self, raw_body: &[u8]) -> Vec<u8>;
}
/// Wraps a function for an endpoint, convertint it to a Handler
struct FnHandler<EP, F>
where
EP: Endpoint,
F: 'static + for<'a> Fn(&'a [u8]) -> <EP as EndpointBody<'a>>::Out,
{
func: F,
_ph: PhantomData<EP>,
}
impl<EP, F> FnHandler<EP, F>
where
EP: Endpoint,
F: 'static + for<'a> Fn(&'a [u8]) -> <EP as EndpointBody<'a>>::Out,
{
pub fn new(func: F) -> Self {
Self {
func,
_ph: PhantomData,
}
}
}
impl<EP, F> Handler for FnHandler<EP, F>
where
EP: Endpoint,
F: 'static + for<'a> Fn(&'a [u8]) -> <EP as EndpointBody<'a>>::Out,
{
fn execute(&self, in_raw_body: &[u8]) -> Vec<u8> {
let body = (self.func)(in_raw_body);
let serialized_body = unimplemented!();
return serialized_body;
}
}
// /////////////////////////////////////////////////////////
/// Collection of handlers
struct Handlers(Vec<Box<dyn Handler>>);
impl Handlers {
pub fn new() -> Self {
Self(vec![])
}
pub fn handle<EP: 'static, F>(&mut self, func: F)
where
EP: Endpoint,
F: 'static + for<'a> Fn(&'a [u8]) -> <EP as EndpointBody<'a>>::Out,
{
self.0.push(Box::new(FnHandler::<EP, F>::new(func)));
}
}
// /////////////////////////////////////////////////////////
struct MyEndpoint;
struct MyEndpointBody<'a> {
pub string: &'a str,
}
impl Endpoint for MyEndpoint {}
impl<'a> EndpointBody<'a> for MyEndpoint {
type Out = MyEndpointBody<'a>;
fn serialize(body: &Self::Out) -> Vec<u8> {
unimplemented!()
}
fn deserialize(raw_body: &'a [u8]) -> Self::Out {
unimplemented!()
}
}
// /////////////////////////////////////////////////////////
fn main() {
let mut handlers = Handlers::new();
handlers.handle::<MyEndpoint, _>(|_body| MyEndpointBody {
string: "test string",
});
handlers.0[1].execute(&[]);
}
我认为应该可以,但是当我检查它时,出现类型错误:
error[E0271]: type mismatch resolving `for<'a> <[closure@src/main.rs:92:38: 94:6] as std::ops::FnOnce<(&'a [u8],)>>::Output == <MyEndpoint as EndpointBody<'a>>::Out`
--> src/main.rs:92:14
|
92 | handlers.handle::<MyEndpoint, _>(|_body| MyEndpointBody {
| ^^^^^^ expected struct `MyEndpointBody`, found associated type
|
= note: expected struct `MyEndpointBody<'_>`
found associated type `<MyEndpoint as EndpointBody<'_>>::Out`
= note: consider constraining the associated type `<MyEndpoint as EndpointBody<'_>>::Out` to `MyEndpointBody<'_>`
= note: for more information, visit https://doc.rust-lang.org/book/ch19-03-advanced-traits.html
这令人困惑,因为MyEndpoint::Out
是MyEndpointBody
,我是从闭包中返回的,但是Rust认为它们不是同一类型。我猜是因为Rust为MyEndpointBody
类型选择了不兼容的匿名生存期,但是我不知道如何解决这个问题。
如何使此代码起作用,以便可以使用具有HRTB关联类型的闭包?
答案 0 :(得分:4)
使用闭包将返回类型包装为新类型可解决此问题:
#![allow(unreachable_code)]
use std::marker::PhantomData;
trait Endpoint: for<'a> EndpointBody<'a> {}
trait EndpointBody<'a> {
type Out: 'a;
fn serialize(body: &Self::Out) -> Vec<u8>;
fn deserialize(raw_body: &'a [u8]) -> Self::Out;
}
struct EPOut<'a, EP: Endpoint>(<EP as EndpointBody<'a>>::Out);
// /////////////////////////////////////////////////////////
/// Trait object compatible handler
trait Handler {
fn execute(&self, raw_body: &[u8]) -> Vec<u8>;
}
/// Wraps a function for an endpoint, convertint it to a Handler
struct FnHandler<EP, F>
where
EP: Endpoint,
F: 'static + for<'a> Fn(&'a [u8]) -> EPOut<'a, EP>,
{
func: F,
_ph: PhantomData<EP>,
}
impl<EP, F> FnHandler<EP, F>
where
EP: Endpoint,
F: 'static + for<'a> Fn(&'a [u8]) -> EPOut<'a, EP>,
{
pub fn new(func: F) -> Self {
Self {
func,
_ph: PhantomData,
}
}
}
impl<EP, F> Handler for FnHandler<EP, F>
where
EP: Endpoint,
F: 'static + for<'a> Fn(&'a [u8]) -> EPOut<'a, EP>,
{
fn execute(&self, in_raw_body: &[u8]) -> Vec<u8> {
let body = (self.func)(in_raw_body);
let serialized_body = unimplemented!();
return serialized_body;
}
}
// /////////////////////////////////////////////////////////
/// Collection of handlers
struct Handlers(Vec<Box<dyn Handler>>);
impl Handlers {
pub fn new() -> Self {
Self(vec![])
}
pub fn handle<EP: 'static, F>(&mut self, func: F)
where
EP: Endpoint,
F: 'static + for<'a> Fn(&'a [u8]) -> EPOut<'a, EP>,
{
self.0.push(Box::new(FnHandler::<EP, F>::new(func)));
}
}
// /////////////////////////////////////////////////////////
struct MyEndpoint;
struct MyEndpointBody<'a> {
pub string: &'a str,
}
impl Endpoint for MyEndpoint {}
impl<'a> EndpointBody<'a> for MyEndpoint {
type Out = MyEndpointBody<'a>;
fn serialize(body: &Self::Out) -> Vec<u8> {
unimplemented!()
}
fn deserialize(raw_body: &'a [u8]) -> Self::Out {
unimplemented!()
}
}
// /////////////////////////////////////////////////////////
fn main() {
let mut handlers = Handlers::new();
handlers.handle::<MyEndpoint, _>(|_body| EPOut(MyEndpointBody {
string: "test string",
}));
handlers.0[1].execute(&[]);
}
考虑到newtype应该与关联的类型几乎相同,因此我很想说这是Rust编译器的错误。似乎还有一些与使用HRTB相关类型有关的ICE:https://github.com/rust-lang/rust/issues/62529
答案 1 :(得分:0)
请问您可以检查that
trait Endpoint: for<'a> DeserializeBody<'a> {}
trait DeserializeBody<'a> {
type Out: 'a;
fn deserialize(raw_body: &'a [u8]) -> Self::Out;
}
fn store_ep<'a, EP, F>(func: F)
where
EP: Endpoint,
F: 'static + Fn(&'a [u8]) -> <EP as DeserializeBody<'a>>::Out,
{
let _ = Box::new(func);
unimplemented!();
}
// /////////////////////////////////////////////////////////
struct MyEndpoint;
struct MyEndpointBody<'a> {
pub string: &'a str,
}
impl Endpoint for MyEndpoint {}
impl<'a> DeserializeBody<'a> for MyEndpoint {
type Out = MyEndpointBody<'a>;
fn deserialize(raw_body: &'a [u8]) -> Self::Out {
unimplemented!();
}
}
// /////////////////////////////////////////////////////////
fn main() {
store_ep::<MyEndpoint, _>(|raw_body| MyEndpointBody { string: "test" });
}
答案 2 :(得分:0)
将DeserializeBody
定义为:
trait DeserializeBody {
type Out;
fn deserialize(raw_body: &[u8]) -> Self::Out;
}
Out
是泛型类型的声明。不要在此处声明生命周期限制,它将在定义站点上明确显示。
这时,Endpoint
的更高等级特质绑定不再必要:
trait Endpoint: DeserializeBody {}
trait DeserializeBody {
type Out;
fn deserialize(raw_body: &[u8]) -> Self::Out;
}
在定义站点,必须为关联的类型Out
表达生存期要求。
如果DeserializeBody
不再是通用名称,则MyEndpoint
必须是:
impl<'a> DeserializeBody for MyEndpoint<'a> {
type Out = MyEndpointBody<'a>;
...
要实现这种要求,我们可以求助于需要寿命'a
的幻像类型。
将所有部分放在一起:
use core::marker::PhantomData;
trait Endpoint: DeserializeBody {}
trait DeserializeBody {
type Out;
fn deserialize(raw_body: &[u8]) -> Self::Out;
}
fn store_ep<EP, F>(func: F)
where
EP: Endpoint,
F: 'static + for<'a> Fn(&'a [u8]) -> <EP as DeserializeBody>::Out,
{
let _ = Box::new(func);
unimplemented!();
}
struct MyEndpoint<'a> {
phantom: PhantomData<&'a ()>
}
struct MyEndpointBody<'a> {
pub string: &'a str,
}
impl<'a> Endpoint for MyEndpoint<'a> {}
impl<'a> DeserializeBody for MyEndpoint<'a> {
type Out = MyEndpointBody<'a>;
fn deserialize(raw_body: &[u8]) -> Self::Out {
unimplemented!();
}
}
fn main() {
store_ep::<MyEndpoint, _>(|raw_body| MyEndpointBody { string: "test" });
}
答案 3 :(得分:0)
我认为问题是您要求处理程序能够使用HK约束处理所有个可能的生命周期-编译器无法证明该生命周期已得到验证,因此无法进行等效处理{ {1}}。
相反,如果您将处理程序参数化为一个唯一的生存期,则似乎可以根据需要进行编译(playground link):
System.Threading.SynchronizationContext.Current