考虑下面的代码 - 编译并运行正常:
use std::rc::Rc;
use std::cell::RefCell;
use crate::Foo::{Something, Nothing};
enum Foo {
Nothing,
Something(i32),
}
fn main() {
let wrapped = Rc::new(RefCell::new(Foo::Nothing));
//....
match *wrapped.borrow() {
Something(x) => println!("{}", x),
Nothing => println!("Nothing"),
};
}
现在我想要匹配两个包装过的值,而不仅仅是一个:
use std::rc::Rc;
use std::cell::RefCell;
use crate::Foo::{Something, Nothing};
enum Foo {
Nothing,
Something(i32),
}
fn main() {
let wrapped1 = Rc::new(RefCell::new(Foo::Nothing));
let wrapped2 = Rc::new(RefCell::new(Foo::Nothing));
//....
match (*wrapped1.borrow(), *wrapped2.borrow()) {
(Something(x), Something(y)) => println!("{}, {}", x, y),
_ => println!("Nothing"),
};
}
现在这会导致编译错误:
error[E0507]: cannot move out of dereference of `std::cell::Ref<'_, Foo>`
--> src\main.rs:16:12
|
16 | match (*wrapped1.borrow(), *wrapped2.borrow()) {
| ^^^^^^^^^^^^^^^^^^ move occurs because value has type `Foo`, which does not implement the `Copy` trait
error[E0507]: cannot move out of dereference of `std::cell::Ref<'_, Foo>`
--> src\main.rs:16:32
|
16 | match (*wrapped1.borrow(), *wrapped2.borrow()) {
| ^^^^^^^^^^^^^^^^^^ move occurs because value has type `Foo`, which does not implement the `Copy` trait
我不太理解这两个示例的语义基本区别。为什么会发生这种情况,如何才能使第二个代码片段正常工作?
&*x.borrow()
语法,Ref<T>
只存在于匹配范围内,因此在match
后稍后安全地获取可变引用(将Ref
保存在变量中需要显式丢弃)。 - vincent