types - 我如何在 Rust 中获得 `std::any::TypeId::of` 超过 `T` 、 `&T` 和 `&mut` T 的相同结果？

Question

我想获取给定类型 T 的一些 TypeId 实例，它实现了 Sized + 'static。

use std::any::TypeId;

trait Comp: Sized + 'static { }

impl<T: 'static> Comp for T { }

struct Foo;

fn main() {
    println!("{}", TypeId::of::<Foo>());
    println!("{}", TypeId::of::<&Foo>());
    println!("{}", TypeId::of::<&mut Foo>());
}

但结果不同。是的，我知道这是完全正常的行为。为了处理这个问题，我修改了这段代码如下。

use std::ops::Deref;

// Same codes are omitted...

fn main() {
    println!("{}", TypeId::of::<Foo>());
    println!("{}", TypeId::of::<<&Foo as Deref>::Target>());
    println!("{}", TypeId::of::<<&mut Foo as Deref>::Target>());
}

现在它打印相同的值。要为所有 Comp 实现此行为，我在下面写道。

use std::any::TypeId;
use std::ops::Deref;

trait Comp: Sized + 'static {
   type Pure: Comp;
}

impl<T: 'static> Comp for T {
   type Pure = T;
}

impl<T: Deref + 'static> Comp for T {
    type Pure = <T as Deref>::Target;
}

struct Foo;

fn main() {
    println!("{}", TypeId::of::<<Foo as Comp>::Pure>());
    println!("{}", TypeId::of::<<&Foo as Comp>::Pure>());
    println!("{}", TypeId::of::<<&mut Foo as Comp>::Pure>());
}

它不会编译，因为我提供了相互冲突的实现。我该如何解决这个问题？

Answer 1

终于在Rust的官方社区找到了解决方案。它可以通过 specialization unstable 特性来实现。

https://play.rust-lang.org/?version=nightly&mode=debug&edition=2018&gist=4c5f206ef3a7843ed57c256ee73ac58c

#![feature(specialization)]
use std::any::TypeId;
use std::ops::Deref;

trait Comp: Sized {
   type Pure: Comp;
}

impl<T> Comp for T {
   default type Pure = T;
}

impl<T: Deref> Comp for T where <T as Deref>::Target: Comp {
    type Pure = <T as Deref>::Target;
}

struct Foo;

fn main() {
    println!("{:?}", TypeId::of::<<Foo as Comp>::Pure>());
    println!("{:?}", TypeId::of::<<&Foo as Comp>::Pure>());
    println!("{:?}", TypeId::of::<<&mut Foo as Comp>::Pure>());
}