rust - 适用于Any的Rust管道运算符(通过重载运算符)

Question

Composition operator and pipe forward operator in Rust

use std::ops::Shr;

struct Wrapped<T>(T);

impl<A, B, F> Shr<F> for Wrapped<A>
where
    F: FnOnce(A) -> B,
{
    type Output = Wrapped<B>;

    fn shr(self, f: F) -> Wrapped<B> {
        Wrapped(f(self.0))
    }
}

fn main() {
    let string = Wrapped(1) >> (|x| x + 1) >> (|x| 2 * x) >> (|x: i32| x.to_string());
    println!("{}", string.0);
}
// prints `4`

这是用于struct的管道运算符的代码:通过重载运算符 Wrapped，但是我需要一个可用于 native 值(我认为是 &dyn Any)的代码。
由于我还不了解Rust的类型系统，所以我喜欢

use std::any::Any;

impl<A, B, F: Fn(A) -> B> Shr<F> for &dyn Any {
    type Output = &dyn Any;

    fn shr(self, f: F) -> &dyn Any {
        f(self.0)
    }
}

但是有明显的错误。
我该如何解决呢？谢谢。

Answer 1

感谢@mcarton，
Rust pipeline operator for Any by overloading operators
我找到了类似的解决方案:
https://docs.rs/apply/0.3.0/apply/trait.Apply.html

pub trait Apply<Res> {
    /// Apply a function which takes the parameter by value.
    fn apply<F: FnOnce(Self) -> Res>(self, f: F) -> Res
    where
        Self: Sized,
    {
        f(self)
    }

    /// Apply a function which takes the parameter by reference.
    fn apply_ref<F: FnOnce(&Self) -> Res>(&self, f: F) -> Res {
        f(self)
    }

    /// Apply a function which takes the parameter by mutable reference.
    fn apply_mut<F: FnOnce(&mut Self) -> Res>(&mut self, f: F) -> Res {
        f(self)
    }
}

impl<T: ?Sized, Res> Apply<Res> for T {
    // use default definitions...
}

fn main() {
    let string = 1
        .apply(|x| x * 2)
        .apply(|x| x + 1)
        .apply(|x: i32| x.to_string());
    println!("{}", string);
}