为什么Deref :: deref本身的返回类型是引用? [英] Why is the return type of Deref::deref itself a reference?

问题描述

我正在阅读Rust的 Deref 特质:

pub trait Deref {
    type Target: ?Sized;
    fn deref(&self) -> &Self::Target;
}

deref函数的类型签名对我来说似乎违反直觉.为什么返回类型是引用?如果引用实现了此特征，因此可以将其取消引用，那么这将产生什么影响?

我能提出的唯一解释是，引用未实现Deref，但被认为是本来可取消引用的".但是，如何编写适用于任何可引用类型(包括Deref<T>和&T)的多态函数呢?

解决方案

引用未实现Deref

的

您可以看到实现Deref的所有类型，并且&T在该列表中:

impl<'a, T> Deref for &'a T where T: ?Sized

不明显的是，当您将*运算符与实现Deref的东西一起使用时，会应用语法糖.看看这个小例子:

use std::ops::Deref;

fn main() {
    let s: String = "hello".into();
    let _: () = Deref::deref(&s);
    let _: () = *s;
}

 error[E0308]: mismatched types
 --> src/main.rs:5:17
  |
5 |     let _: () = Deref::deref(&s);
  |                 ^^^^^^^^^^^^^^^^ expected (), found &str
  |
  = note: expected type `()`
             found type `&str`

error[E0308]: mismatched types
 --> src/main.rs:6:17
  |
6 |     let _: () = *s;
  |                 ^^ expected (), found str
  |
  = note: expected type `()`
             found type `str`
 

对deref的显式调用返回&str，但是运算符*返回str.就像您正在调用*Deref::deref(&s)一样，忽略了隐式的无限递归.

Xirdus说的是正确的

如果deref返回一个值，它要么会因为总是移出而变得无用，要么具有与其他函数完全不同的语义

尽管没用"有点强；对于实现Copy的类型仍然有用.

另请参阅:

• 为什么对Deref :: deref的结果进行断言会因类型不匹配而失败?

请注意，以上所有内容对于Index和IndexMut都是有效的.

I was reading the docs for Rust's Deref trait:

pub trait Deref {
    type Target: ?Sized;
    fn deref(&self) -> &Self::Target;
}

The type signature for the deref function seems counter-intuitive to me; why is the return type a reference? If references implement this trait so they can be dereferenced, what effect would this have at all?

The only explanation that I can come up with is that references don't implement Deref, but are considered "primitively dereferenceable". However, how would a polymorphic function which would work for any dereferenceable type, including both Deref<T> and &T, be written then?

解决方案

that references don't implement Deref

You can see all the types that implement Deref, and &T is in that list:

impl<'a, T> Deref for &'a T where T: ?Sized

The non-obvious thing is that there is syntactical sugar being applied when you use the * operator with something that implements Deref. Check out this small example:

use std::ops::Deref;

fn main() {
    let s: String = "hello".into();
    let _: () = Deref::deref(&s);
    let _: () = *s;
}

error[E0308]: mismatched types
 --> src/main.rs:5:17
  |
5 |     let _: () = Deref::deref(&s);
  |                 ^^^^^^^^^^^^^^^^ expected (), found &str
  |
  = note: expected type `()`
             found type `&str`

error[E0308]: mismatched types
 --> src/main.rs:6:17
  |
6 |     let _: () = *s;
  |                 ^^ expected (), found str
  |
  = note: expected type `()`
             found type `str`

The explicit call to deref returns a &str, but the operator * returns a str. It's more like you are calling *Deref::deref(&s), ignoring the implied infinite recursion.

Xirdus is correct in saying

If deref returned a value, it would either be useless because it would always move out, or have semantics that drastically differ from every other function

Although "useless" is a bit strong; it would still be useful for types that implement Copy.

See also:

• Why does asserting on the result of Deref::deref fail with a type mismatch?

Note that all of the above is effectively true for Index and IndexMut as well.

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