我正在尝试为枚举实现Deref:
use std::rc::Rc;
use std::ops::Deref;
pub trait tObject {
fn name(&self) -> String;
fn span(&self) -> u32;
}
pub struct t1 {
pub name: String,
pub bd: Option<String>,
pub span: u32,
pub label: Option<String>
}
pub struct t2 {
pub name: String,
pub vrf: Option<String>,
pub span: u32,
pub label: Option<String>,
pub svi: u32
}
pub struct t3 {
pub name: String,
pub span: u32,
pub label: Option<String>
}
impl tObject for t1 {
fn name(&self) -> String {self.name.clone()}
fn span(&self) -> u32 {self.span.clone()}
}
impl tObject for t2 {
fn name(&self) -> String {self.name.clone()}
fn span(&self) -> u32 {self.span.clone()}
}
impl tObject for t3 {
fn name(&self) -> String {self.name.clone()}
fn span(&self) -> u32 {self.span.clone()}
}
pub enum TType {
t1(Rc<t1>),
t2(Rc<t2>),
t3(Rc<t3>)
}
impl Deref for TType {
type Target = tObject;
fn deref<'a>(&'a self) -> &'a tObject {
match *self {
TType::t1(ref x) => x as &t1,
TType::t2(ref x) => x as &t2,
TType::t3(ref x) => x as &t3
}
}
}
fn main() {
let mut t1s: Vec<Rc<t1>> = Vec::new();
let mut t2s: Vec<Rc<t2>> = Vec::new();
let mut t3s: Vec<Rc<t3>> = Vec::new();
let t_iter: Box<Iterator<Item=TType>> = Box::new(t1s.iter().map(|x| TType::t1(x.clone())).chain(
t2s.iter().map(|x| TType::t2(x.clone())).chain(
t3s.iter().map(|x| TType::t3(x.clone())))));
}
编译器报错:
rustc 1.15.1 (021bd294c 2017-02-08)
error[E0495]: cannot infer an appropriate lifetime for lifetime parameter 'a in generic type due to conflicting requirements
--> <anon>:53:5
|
53 | fn deref<'a>(&'a self) -> &'a tObject {
| _____^ starting here...
54 | | match *self {
55 | | TType::t1(ref x) => x as &t1,
56 | | TType::t2(ref x) => x as &t2,
57 | | TType::t3(ref x) => x as &t3
58 | | }
59 | | }
| |_____^ ...ending here
|
note: first, the lifetime cannot outlive the anonymous lifetime #1 defined on the body at 53:42...
--> <anon>:53:43
|
53 | fn deref<'a>(&'a self) -> &'a tObject {
| ___________________________________________^ starting here...
54 | | match *self {
55 | | TType::t1(ref x) => x as &t1,
56 | | TType::t2(ref x) => x as &t2,
57 | | TType::t3(ref x) => x as &t3
58 | | }
59 | | }
| |_____^ ...ending here
note: ...so that method type is compatible with trait (expected fn(&TType) -> &tObject + 'static, found fn(&TType) -> &tObject)
--> <anon>:53:5
|
53 | fn deref<'a>(&'a self) -> &'a tObject {
| _____^ starting here...
54 | | match *self {
55 | | TType::t1(ref x) => x as &t1,
56 | | TType::t2(ref x) => x as &t2,
57 | | TType::t3(ref x) => x as &t3
58 | | }
59 | | }
| |_____^ ...ending here
= note: but, the lifetime must be valid for the static lifetime...
note: ...so that method type is compatible with trait (expected fn(&TType) -> &tObject + 'static, found fn(&TType) -> &tObject)
--> <anon>:53:5
|
53 | fn deref<'a>(&'a self) -> &'a tObject {
| _____^ starting here...
54 | | match *self {
55 | | TType::t1(ref x) => x as &t1,
56 | | TType::t2(ref x) => x as &t2,
57 | | TType::t3(ref x) => x as &t3
58 | | }
59 | | }
| |_____^ ...ending here
如果我将返回类型设为 deref Self::Target 而不是 tObject，它可以正常编译。我不明白这种行为。
最佳答案
这是一个MCVE .程序员使用这些来帮助缩小问题的范围。例如，此 MCVE 排除了由于使用 enum、Rc、特征中的任何方法或结构中的任何字段而导致的任何情况.这使我们能够专注于重要的事情:
use std::ops::Deref;
pub trait Trait {}
pub struct S {}
impl Trait for S {}
pub struct Container(S);
impl Deref for Container {
type Target = Trait;
// fn deref(&self) -> &Trait { // Fails!
fn deref(&self) -> &Self::Target { // Works!
&self.0
}
}
从代码中，我们可以凭直觉知道，不知何故，Trait 和 Self::Target 不是 相同的类型。在这里查看类型有点棘手，但是这段代码将类型打印为编译器错误:
fn deref(&self) -> &Self::Target {
let a: Self::Target;
&self.0
}
error[E0277]: the trait bound `Trait + 'static: std::marker::Sized` is not satisfied
我们实际上并不关心错误，但我们发现了类型:Trait + 'static。让我们看看如果我们尝试类似的事情会发生什么:
fn deref(&self) -> &(Trait + 'static) {
&self.0
}
这编译。
如果您不熟悉此语法，有很多关于它的问题。这里有一些:
关于rust - 实现 Deref 特征时无法为生命周期参数推断出合适的生命周期，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42309597/