在 Rust 中设计一个具有可变 getter 和不可变 getter 的特征，并默认实现不可变的 getter

不，这不合理。

用户可以在其实现中修改 ，这也将修改 中的结构。selfinside_mutinside_ref

喜欢这个：

pub trait MyTrait<T> {
    fn inside_mut(&mut self) -> &mut T;

    fn inside_ref(&self) -> &T {
        let ptr = self as *const Self as *mut Self;
        &*unsafe { (&mut *ptr).inside_mut() }
    }
}

#[derive(Debug)]
struct MyStruct {
    value: i32,
}

impl MyTrait<i32> for MyStruct {
    fn inside_mut(&mut self) -> &mut i32 {
        self.value += 1;
        &mut self.value
    }
}

fn main() {
    // Important: this is not `mut`!!!
    let x = MyStruct { value: 42 };

    println!("{:?}", x);
    x.inside_ref();
    println!("{:?}", x);
}

MyStruct { value: 42 }
MyStruct { value: 43 }

这意味着这个实现是不合理的：

被视为未定义的行为

• 更改不可变数据。项目内的所有数据都是不可变的。此外，通过共享引用或不可变绑定拥有的数据访问的所有数据都是不可变的，除非该数据包含在 .constUnsafeCell<U>

在看了Finomnis的回答之后，我来到了这个解决方案（playground），miri不会产生UB警报：

use core::cell::UnsafeCell;
pub trait MyTrait<T> {
    fn inside_mut(&mut self) -> &mut T {
        let ptr = self.implement_ref().get();
        unsafe { &mut *ptr }
    }

    fn inside_ref(&self) -> &T {
        let ptr = self.implement_ref().get();
        unsafe { & *ptr }
    }

    fn implement_ref(&self) -> &UnsafeCell<T>;
}

pub struct Data(UnsafeCell<String>);

impl Data {
    pub fn new(s:String) -> Self {
        Data(UnsafeCell::new(s))
    }    
}

impl MyTrait<String> for Data {
    fn implement_ref(&self) -> &UnsafeCell<String> { &self.0 }
}

fn main() {
    let mut data = Data::new(format!("Foo"));
    let x: &String = data.inside_ref();
    let y: &String = data.inside_ref();
    println!("x -> {x}");
    println!("y -> {y}");
    *data.inside_mut() = format!("Bar");
    println!("data.0 -> {}", data.inside_ref());
}

但是，我仍然有一个问题。miri 为以下代码 （playground） 生成 UB 警报：

pub trait MyTrait<T> {
    fn inside_ref(&self) -> &T;
    
    fn inside_mut(&mut self) -> &mut T  {
        let ptr = self.inside_ref() as *const T as *mut T;
        unsafe { &mut *ptr }
    }
}

pub struct Data(String);
impl MyTrait<String> for Data {
    fn inside_ref(&self) -> &String { &self.0 }
}

fn main() {
    let mut data = Data(format!("Foo"));
    let x: &String = data.inside_ref();
    let y: &String = data.inside_ref();
    println!("x -> {x}");
    println!("y -> {y}");
    *data.inside_mut() = format!("Bar");
    println!("data.0 -> {}", data.0);
}

但是 Finomnis 提供的很好的反例不适用于这种情况。有人能举例说明后一种情况的不合理性吗？

编辑：灵感来自cafce25（playground）的反例：

static IMMUTABLE_STATIC_STR: &'static str = "IMMUTABLE FOO";

pub trait MyTrait<T> {
    fn inside_ref(&self) -> &T;
    
    fn inside_mut(&mut self) -> &mut T  {
        let ptr = self.inside_ref() as *const T as *mut T;
        unsafe { &mut *ptr }
    }
}

pub struct Data;
impl MyTrait<&'static str> for Data {
    fn inside_ref(&self) -> &&'static str { &IMMUTABLE_STATIC_STR }
}

fn main() {
    let mut data = Data;
    let x: &&'static str = data.inside_ref();
    let y: &&'static str = data.inside_ref();
    println!("x -> {x}");
    println!("y -> {y}");
    *data.inside_mut() = "Bar";
    println!("data.inside_ref() -> {}", data.inside_ref());
}

结果：

Standard Error

   Compiling playground v0.0.1 (/playground)
    Finished release [optimized] target(s) in 1.61s
     Running `target/release/playground`

Standard Output

x -> IMMUTABLE FOO
y -> IMMUTABLE FOO
data.inside_ref() -> IMMUTABLE FOO