Question

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())))));
}

编译器报告此错误：

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，则编译正常。我不明白这种行为。

Answer 1

这是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事情。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，我们可以直觉地知道，Trait和Self::Target 不是相同的类型。看到这里的类型有点棘手，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这样的事情会发生什么：

fn deref(&self) -> &(Trait + 'static) {
    &self.0
}

编译。

如果您不熟悉此语法，则有很多问题。以下是一些：

在实现Deref特征时，无法推断生命周期参数的适当生命周期

1 个答案: