refcell rc<refcell<>> 不会改变

Question

这是我的简化代码。我需要更改函数中的 pdp_state。但状态仍然是'a'。我不明白为什么克隆 Rc 不起作用。我还检查了这个克隆的结构，它工作正常。

#[derive(Clone,Copy)]
enum PDPStatus{
    a,
    b
}
struct network{
    pdp_state:Rc<RefCell<PDPStatus>>,
}
impl network{
    fn set(&mut self){
        let mut t = *self.pdp_state.clone().borrow_mut();
        match t {
            a => {let m1 = self.pdp_state.clone();
                let mut a = (*m1).borrow_mut() ;
                *a = PDPStatus::b;
                println!("a");},
            b=> {let m1 = self.pdp_state.clone();m1.replace( PDPStatus::a);
                println!("b");},
        };

    }
}
fn main() {
    let mut network1 = network::new();
    network1.set();
    network1.set();
    network1.set();
    network1.set();
}

更新： 我的 set 函数看起来像这样。我需要两个可以访问 pdp_state 的闭包。我将这些闭包作为回调传递。我确信这些闭包不会一起调用。

    fn set(&mut self){
        let borrowed_pdp_status = self.pdp_state.borrow().clone();
        match borrowed_pdp_status {
            PDPStatus::a => {
                let mut state = self.pdp_state.clone();
                let mut closuree = || state =  Rc::new(RefCell::new(PDPStatus::b));
                let mut state1 = self.pdp_state.clone();
                let mut closuree1 = || state1 = Rc::new(RefCell::new(PDPStatus::b));

                closuree();
                closuree1();
                println!("a");
            },
            PDPStatus::b => {
                let mut closuree = || self.pdp_state = Rc::new(RefCell::new(PDPStatus::a));
                closuree();
                println!("b");
            },
        };
    }

Answer 1

在set方法中，需要先borrowself.pdp_state再clone()在一个变量中，然后匹配你克隆的那个变量。

将set 方法替换为：

fn set(&mut self) {
        let borrowed_pdp_status = self.pdp_state.borrow().clone();
        match borrowed_pdp_status {
            PDPStatus::a => {
                self.pdp_state = Rc::new(RefCell::new(PDPStatus::b));
                println!("a");
            },
            PDPStatus::b => {
                self.pdp_state = Rc::new(RefCell::new(PDPStatus::a));
                println!("b");
            },
        };
}

游乐场链接 - https://play.rust-lang.org/?version=stable&mode=release&edition=2018&gist=426d4cb7db9a92ee4ddcd4f36dbc12f7

Answer 2

此答案是在问题更新后发布的：

编辑

您可以使用RefCell 中的replace() 方法

fn set(&mut self) {
    let borrowed_pdp_status = self.pdp_state.borrow().clone();
    match borrowed_pdp_status {
        PDPStatus::a => {
            let mut closuree = || {
                self.pdp_state.replace(PDPStatus::b);
            };
            let mut closuree1 = || {
                self.pdp_state.replace(PDPStatus::b);
            };

            closuree();
            closuree1();
            println!("a");
        }
        PDPStatus::b => {
            let mut closuree = || {
                self.pdp_state.replace(PDPStatus::a);
            };
            closuree();
            println!("b");
        }
    };
}

游乐场链接 - https://play.rust-lang.org/?version=stable&mode=release&edition=2018&gist=4af02228d58b2f2c865a525e3f70d6a0

旧

您可以让闭包将&mut self.pdp_state 作为参数，然后对其进行更新。

fn set(&mut self) {
    let borrowed_pdp_status = self.pdp_state.borrow().clone();
    match borrowed_pdp_status {
        PDPStatus::a => {
            let mut closuree = |local_pdp_state: &mut Rc<RefCell<PDPStatus>>| {
                *local_pdp_state = Rc::new(RefCell::new(PDPStatus::b))
            };
            let mut closuree1 = |local_pdp_state: &mut Rc<RefCell<PDPStatus>>| {
                *local_pdp_state = Rc::new(RefCell::new(PDPStatus::b))
            };

            closuree(&mut self.pdp_state);
            closuree1(&mut self.pdp_state);
            println!("a");
        }
        PDPStatus::b => {
            let mut closuree = |local_pdp_state: &mut Rc<RefCell<PDPStatus>>| {
                *local_pdp_state = Rc::new(RefCell::new(PDPStatus::a))
            };
            closuree(&mut self.pdp_state);
            println!("b");
        }
    };
}

游乐场链接 - https://play.rust-lang.org/?version=stable&mode=release&edition=2018&gist=4af96385b0446082afdb7d615bb8eecb