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我有一个需要变量所有权的第三方库的功能。不幸的是，这个变量在一个Rc<RefCell<Option<Foo>>>.

我的代码看起来像这样简化：

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

pub struct Foo {
    val: i32,
}

fn main() {
    let foo: Rc<RefCell<Option<Foo>>> = Rc::new(RefCell::new(Some(Foo { val: 1 })));

    if let Some(f) = foo.into_inner() {
        consume_foo(f);
    }
}

fn consume_foo(f: Foo) {
    println!("Foo {} consumed", f.val)
}

error[E0507]: cannot move out of an `Rc`
  --> src/main.rs:11:22
   |
11 |     if let Some(f) = foo.into_inner() {
   |                      ^^^ move occurs because value has type `std::cell::RefCell<std::option::Option<Foo>>`, which does not implement the `Copy` trait

我尝试使用std::mem::replace(...)如何在对结构的可变引用中为字段交换新值？：

fn main() {
    let mut foo: Rc<RefCell<Option<Foo>>> = Rc::new(RefCell::new(Some(Foo { val: 1 })));

    let mut foo_replaced = std::mem::replace(&mut foo.into_inner(), None);
    if let Some(f) = foo_replaced.take() {
        consume_foo(f);
    }
}

error[E0507]: cannot move out of an `Rc`
  --> src/main.rs:11:51
   |
11 |     let mut foo_replaced = std::mem::replace(&mut foo.into_inner(), None);
   |                                                   ^^^ move occurs because value has type `std::cell::RefCell<std::option::Option<Foo>>`, which does not implement the `Copy` trait

我不知道如何正确地做到这一点。

4

0 回答 0