7

例如给出这段代码： 

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

// Don't want to copy for performance reasons
struct LibraryData {
    // Fields ...
}

// Creates and mutates data field in methods
struct LibraryStruct {
    // Only LibraryStruct should have mutable access to this
    data: Rc<RefCell<LibraryData>>
}

impl LibraryStruct {
    pub fn data(&self) -> Rc<RefCell<LibraryData>> {
        self.data.clone()
    }
}

// Receives data field from LibraryStruct.data()
struct A {
    data: Rc<RefCell<LibraryData>>
}

impl A {
    pub fn do_something(&self) {
        // Do something with self.data immutably

        // I want to prevent this because it can break LibraryStruct
        // Only LibraryStruct should have mutable access 
        let data = self.data.borrow_mut();
        // Manipulate data
    }
}

如何防止LibraryData在外部发生变异LibraryStruct？LibraryStruct应该是唯一能够data在其方法中发生变异的人。这是可能的Rc<RefCell<LibraryData>>还是有替代方案？请注意，我正在编写“库”代码，以便可以更改它。

4

1 回答 1

8

如果你分享 aRefCell那么它总是有可能改变它 - 这基本上就是它的全部意义。鉴于您能够更改 的实现LibraryStruct，您可以确保它data不是公开的，并通过 getter 方法控制它如何向用户公开：

pub struct LibraryStruct {
    // note: not pub
    data: Rc<RefCell<LibraryData>>
}

impl LibraryStruct {
    // could also have returned `Ref<'a, LibraryData> but this hides your 
    // implementation better
    pub fn data<'a>(&'a self) -> impl Deref<Target = LibraryData> + 'a {
        self.data.borrow()
    }
}

在您的其他结构中，您可以通过将其视为参考来保持简单：

pub struct A<'a> {
    data: &'a LibraryData,
}

impl<'a> A<'a> {
    pub fn do_something(&self) {
        // self.data is only available immutably here because it's just a reference
    }
}

fn main() { 
    let ld = LibraryData {};
    let ls = LibraryStruct { data: Rc::new(RefCell::new(ld)) };

    let a = A { data: &ls.data() };
}

如果您需要更长时间地保留引用，在此期间RefCell需要在库代码中可变地借用原始引用，那么您需要制作一个可以管理它的自定义包装器。这可能有一个标准库类型，但我不知道，而且很容易为您的用例专门制作一些东西：

// Wrapper to manage a RC<RefCell> and make it immutably borrowable
pub struct ReadOnly<T> {
    // not public
    inner: Rc<RefCell<T>>,
}

impl<T> ReadOnly<T> {
    pub fn borrow<'a>(&'a self) -> impl Deref<Target = T> + 'a {
        self.inner.borrow()
    }
}

现在在您的库代码中返回它：

impl LibraryStruct {
    pub fn data<'a>(&'a self) -> ReadOnly<LibraryData> {
        ReadOnly { inner: self.data.clone() }
    }
}

并且当你使用它时，内部RefCell将无法直接访问，并且数据只能用于不可变地借用：

pub struct A {
    data: ReadOnly<LibraryData>,
}

impl A {
    pub fn do_something(&self) {
        //  data is immutable here
        let data = self.data.borrow();
    }
}

fn main() { 
    let ld = LibraryData {};
    let ls = LibraryStruct { data: Rc::new(RefCell::new(ld)) };

    let a = A { data: ls.data() };
}
于 2018-09-01T21:07:04.043 回答