我如何解决RefCell问题？

Question

我有一个带有两个C指针和一个Rust的结构HashMap.

struct MyStruct {
    p1: *mut ...,
    p2: *mut ...,
    hm: Box<HashMap<...>>
}

我的结构被处理为一个Rc<RefCell<MyStruct>>,我有一个C函数被调用,如下所示:

c_call(my_struct.borrow().p1, my_struct.borrow().p2);

C具有防锈回调被调用时的执行c_call需要一个my_struct.borrow_mut(),但my_struct已经借用c_call其需要p1和p2,所以我得到RefCell<T> already borrowed.

问题是c_call不能改变的,它需要一成不变的访问p1,并p2与一些borrow_mut的my_struct.

这是一个MCVE:

use std::cell::RefCell;
use std::collections::HashMap;
use std::mem::uninitialized;
use std::os::raw::c_void;
use std::rc::Rc;

struct MyStruct {
    p1: *mut c_void,
    p2: *mut c_void,
    hm: Box<HashMap<String, String>>
}

// c_call can't mutate hm because my_struct is already borrowed
// c_call can't be changed
fn c_call(_p1: *mut c_void, _p2: *mut c_void, my_struct: Rc<RefCell<MyStruct>>) {
    my_struct.borrow_mut().hm.insert("hey".to_string(), "you".to_string());
}

// call only receives Rc<RefCell<MyStruct>> and need to call c_call
fn call(my_struct: Rc<RefCell<MyStruct>>) {
    c_call(my_struct.borrow().p1, my_struct.borrow().p2, my_struct.clone());
}

fn main() {
    unsafe {
        let my_struct = MyStruct {
            p1: uninitialized::<*mut c_void>(), // irrelevant
            p2: uninitialized::<*mut c_void>(),
            hm: Box::new(HashMap::new())
        };

        let my_struct = Rc::new(RefCell::new(my_struct));

        call(my_struct);
    }
}

(围栏)

我该如何解决这个问题？

Answer 1

您的问题是调用调用borrow()的参数c_call将借用该对象,直到调用结束.如果你改变它

let (p1, p2) = {
    let x = my_struct.borrow();
    (x.p1, x.p2)
};
c_call(p1, p2, my_struct.clone());

然后借了前结束c_call通话,这样c_call可以borrow_mut你的对象了.