The following未编译:
use std::any::Any;
pub trait CloneBox: Any {
fn clone_box(&self) -> Box<dyn CloneBox>;
}
impl<T> CloneBox for T
where
T: Any + Clone,
{
fn clone_box(&self) -> Box<dyn CloneBox> {
Box::new(self.clone())
}
}
struct Foo(Box<dyn CloneBox>);
impl Clone for Foo {
fn clone(&self) -> Self {
let Foo(b) = self;
Foo(b.clone_box())
}
}
错误消息:
error[E0495]: cannot infer an appropriate lifetime for pattern due to conflicting requirements
--> src/lib.rs:20:17
|
20 | let Foo(b) = self;
| ^
|
note: first, the lifetime cannot outlive the anonymous lifetime #1 defined on the method body at 19:5...
--> src/lib.rs:19:5
|
19 | / fn clone(&self) -> Self {
20 | | let Foo(b) = self;
21 | | Foo(b.clone_box())
22 | | }
| |_____^
note: ...so that reference does not outlive borrowed content
--> src/lib.rs:20:17
|
20 | let Foo(b) = self;
| ^
= note: but, the lifetime must be valid for the static lifetime...
note: ...so that the type `&std::boxed::Box<dyn CloneBox>` will meet its required lifetime bounds
--> src/lib.rs:21:15
|
21 | Foo(b.clone_box())
|
但是,如果将 clone()
中的代码从 Foo(b.clone_box())
更改为 Foo(self.0.clone_box())
,编译没有问题。理论上,字段访问应该和模式匹配一样,但是为什么模式匹配会存在生命周期问题呢?
在我的真实代码中,数据位于枚举中,而不是结构中,因此模式匹配是唯一的选择。
最佳答案
TL;DR:在调用方法之前取消引用该值:
Foo((*b).clone_box())
与 let Foo(b) = self
,类型 b
是 &Box<(dyn CloneBox + 'static)>
。方法调用有效
Foo(<&Box<dyn CloneBox + 'static> as CloneBox>::clone_box(&b))
该值不能成为特征对象Box<dyn CloneBox + 'static>
因为本地引用。有趣的是,我相信如果编译器允许的话,这将递归地使用毯子实现。
与 self.0.clone_box()
,方法调用有效:
Foo(<dyn CloneBox as CloneBox>::clone_box(&**b)
我们可以将其写为 Foo((&**b).clone_box())
明确地说,但由于没有中间实现,Foo((*b).clone_box())
就足够了。
另请参阅:
关于rust - 为什么在解构变量后无法调用方法,但直接访问该字段却可以调用方法?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54757677/