Rust面向对象设计模式

我正试着通过阅读The Rust Programming Language Book来学习《 rust 》.我现在读的是第17章，在第Trade-offs of the State Pattern章的末尾，你们需要做以下事情:

• 需要两次调用才能将状态更改为已发布.

你得到的代码是:

pub struct Post {
    state: Option<Box<dyn State>>,
    content: String
}

impl Post {
    pub fn new() -> Post {
        Post {
            state: Some(Box::new(Draft {})),
            content: String::new(),
        }
    }

    pub fn add_text(&mut self, text: &str) {
        self.content.push_str(text);
    }

    pub fn content(&self) -> &str {
        self.state.as_ref().unwrap().content(self)
    }

    pub fn request_review(&mut self) {
        if let Some(s) = self.state.take() {
            self.state = Some(s.request_review())
        }
    }

    pub fn approve(&mut self) {
        if let Some(s) = self.state.take() {
            self.state = Some(s.approve())
        }
    }
}

trait State {
    fn request_review(self: Box<Self>) -> Box<dyn State>;
    fn approve(self: Box<Self>) -> Box<dyn State>;
    fn content<'a>(&self, post: &'a Post) -> &'a str {
        ""
    }
}

struct Draft {}

impl State for Draft {
    fn request_review(self: Box<Self>) -> Box<dyn State> {
        Box::new(PendingReview {})
    }

    fn approve(self: Box<Self>) -> Box<dyn State> {
        self
    }
}

struct PendingReview {}

impl State for PendingReview {
    fn request_review(self: Box<Self>) -> Box<dyn State> {
        self
    }

    fn approve(self: Box<Self>) -> Box<dyn State> {
        Box::new(Published {})
    }
}

struct Published {}

impl State for Published {
    fn request_review(self: Box<Self>) -> Box<dyn State> {
        self
    }

    fn approve(self: Box<Self>) -> Box<dyn State> {
        self
    }

    fn content<'a>(&self, post: &'a Post) -> &'a str {
        &post.content
    }
}

所以我所理解的是，without breaking the trait signature，你必须实现某种逻辑到PendingReview，才能需要两次批准才能更改为Published状态.

我想做一些像这样的事情

struct PendingReview {
    reviews: i32,
}

impl PendingReview {
    fn add_review(&mut self) {
        self.reviews += 1;
    }

    pub fn ready_to_approve(&self) -> bool {
        if self.reviews == 2 {
            return true;
        }
        add_review();
        return false;
    }
}

然后在PendingReview状态的approve函数中调用ready_to_approve，但这不起作用，因为我需要一个我没有的可变引用，如果不打破特征签名，我就不能拥有它.此外，这种方法可能还有更多我看不到的问题，实际上我对Rust还很陌生.

我真的不知道如何处理这个问题，所以任何帮助都是感激的.