匹配内部可变枚举的预期方法是啥？

Posted 2023-02-19

【中文标题】匹配内部可变枚举的预期方法是啥？

【英文标题】：What is the intended way to match an interior mutable enum?匹配内部可变枚举的预期方法是什么？

【发布时间】：2019-06-26 22:50:28

【问题描述】：

这些是我可以想出的尝试匹配引用计数的内部可变枚举的方法：

#![allow(unused)]

use std::cell::RefCell;
use std::rc::Rc;

#[derive(Debug, Clone, PartialEq)]
struct Bar 
    some_bar: Vec<f64>,

#[derive(Debug, Clone, PartialEq)]
struct Baz 
    some_baz: i32,


#[derive(Debug, Clone, PartialEq)]
enum Foo 
    Bar(Bar),
    Baz(Baz),


fn is_baz(foo_ref: Rc<RefCell<Foo>>) -> bool 
    match foo_ref 
        Foo::Bar(_) => false,
        Foo::Baz(_) => true,
    


fn is_baz_borrow(foo_ref: Rc<RefCell<Foo>>) -> bool 
    match foo_ref.borrow() 
        Foo::Bar(_) => false,
        Foo::Baz(_) => true,
    


fn is_baz_deref(foo_ref: Rc<RefCell<Foo>>) -> bool 
    match *foo_ref 
        Foo::Bar(_) => false,
        Foo::Baz(_) => true,
    


fn is_baz_get_mut(foo_ref: Rc<RefCell<Foo>>) -> bool 
    match foo_ref.get_mut() 
        Foo::Bar(_) => false,
        Foo::Baz(_) => true,
    


fn is_baz_as_ref(foo_ref: Rc<RefCell<Foo>>) -> bool 
    match foo_ref.as_ref() 
        Foo::Bar(_) => false,
        Foo::Baz(_) => true,
    


fn is_baz_as_ref_borrow(foo_ref: Rc<RefCell<Foo>>) -> bool 
    match foo_ref.as_ref().borrow() 
        Foo::Bar(_) => false,
        Foo::Baz(_) => true,
    


fn is_baz_get_mut_mut(mut foo_ref: Rc<RefCell<Foo>>) -> bool 
    match foo_ref.get_mut() 
        Foo::Bar(_) => false,
        Foo::Baz(_) => true,
    


fn main() 
    let foo = Foo::Bar(Bar 
        some_bar: vec![1.1, 999.0],
    );
    let foo_ref = Rc::new(RefCell::new(foo));

    // none of these work
    assert!(is_baz(foo_ref.clone()));
    assert!(is_baz_borrow(foo_ref.clone()));
    assert!(is_baz_deref(foo_ref.clone()));
    assert!(is_baz_get_mut(foo_ref.clone()));
    assert!(is_baz_as_ref(foo_ref.clone()));
    assert!(is_baz_as_ref_borrow(foo_ref.clone()));

    // this works
    assert!(is_baz_get_mut_mut(foo_ref.clone()));

Playground

但是，大多数方式都会产生类型不匹配的错误，例如

error[E0308]: mismatched types
  --> src/main.rs:24:9
   |
24 |         Foo::Baz(_) => true
   |         ^^^^^^^^^^^ expected struct `std::rc::Rc`, found enum `Foo`
   |
   = note: expected type `std::rc::Rc<std::cell::RefCell<Foo>>`
              found type `Foo`

或

error[E0308]: mismatched types
  --> src/main.rs:30:9
   |
30 |         Foo::Bar(_) => false,
   |         ^^^^^^^^^^^ expected struct `std::cell::Ref`, found enum `Foo`
   |
   = note: expected type `std::cell::Ref<'_, Foo, >`
              found type `Foo`

唯一可行的方法是：

match foo_ref.get_mut() 
    Foo::Bar(_) => false,
    Foo::Baz(_) => true

这是预期的方式吗？即使是只读访问？

【参考方案1】：

您可以使用Ref 的Deref impl：

fn is_baz_get(foo_ref: Rc<RefCell<Foo>>) -> bool 
    match *foo_ref.borrow() 
        Foo::Bar(_) => false,
        Foo::Baz(_) => true
    

【讨论】：

非常感谢！这让我发疯了！如果我使用 ref rust fn is_baz_of_5_deref_borrow(foo_ref: Rc<RefCell<Foo>>) -> bool match *foo_ref.borrow() Foo::Bar(_) => false, Foo::Baz(ref baz) => baz ==&Bazsome_baz:5

，我什至可以读取枚举的内容