[英]Cannot infer a lifetime for a closure returning a boxed trait that contains a reference
我正在尝试编译以下代码( Playground ):
trait MockFutureTrait {
type Item;
}
struct MockFuture<T> {
item: T,
}
impl<T> MockFutureTrait for MockFuture<T> {
type Item = T;
}
struct FragMsgReceiver<'a, 'c: 'a> {
recv_dgram: &'a FnMut(&mut [u8])
-> Box<MockFutureTrait<Item = &mut [u8]> + 'c>,
}
fn constrain_handler<F>(f: F) -> F
where
F: FnMut(&mut [u8]) -> Box<MockFutureTrait<Item = &mut [u8]>>,
{
f
}
fn main() {
let mut recv_dgram = constrain_handler(|buf: &mut [u8]| {
Box::new(MockFuture { item: buf }) as Box<MockFutureTrait<Item = &mut [u8]>>
});
let ref_recv_dgram = &mut recv_dgram;
let fmr = FragMsgReceiver {
recv_dgram: ref_recv_dgram,
};
}
我得到了编译错误:
error[E0495]: cannot infer an appropriate lifetime due to conflicting requirements
--> src/main.rs:28:37
|
28 | Box::new(MockFuture { item: buf }) as Box<MockFutureTrait<Item = &mut [u8]>>
| ^^^
|
note: first, the lifetime cannot outlive the anonymous lifetime #2 defined on the body at 27:44...
--> src/main.rs:27:44
|
27 | let mut recv_dgram = constrain_handler(|buf: &mut [u8]| {
| ____________________________________________^
28 | | Box::new(MockFuture { item: buf }) as Box<MockFutureTrait<Item = &mut [u8]>>
29 | | });
| |_____^
note: ...so that expression is assignable (expected &mut [u8], found &mut [u8])
--> src/main.rs:28:37
|
28 | Box::new(MockFuture { item: buf }) as Box<MockFutureTrait<Item = &mut [u8]>>
| ^^^
= note: but, the lifetime must be valid for the static lifetime...
note: ...so that expression is assignable (expected std::boxed::Box<MockFutureTrait<Item=&mut [u8]> + 'static>, found std::boxed::Box<MockFutureTrait<Item=&mut [u8]>>)
--> src/main.rs:28:9
|
28 | Box::new(MockFuture { item: buf }) as Box<MockFutureTrait<Item = &mut [u8]>>
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
我试图添加各种生命周期提示,但是无法编译此代码。
我之前在SO上与此相关的问题:
无法推断出包含对闭包的引用的结构的生存期 :如果返回值是简单结构而不是特征,则可以解决相同的问题。
多个结构字段如何成为使用相同寿命的泛型的泛型? :关于尝试不使用Box来解决此问题。 答案表明,现在我将不得不使用Box>。
注意,我根据问题2的建议使用了辅助函数constrain_handler
。 它使我能够克服一个不同的编译错误。
您似乎错过了先前问题及其重复内容的关键要点:
通过在闭包参数上声明类型,可以停止对参数执行类型推断 。 这将导致闭包生成新的隐式生存期,这与您的要求不符。 只是根本不声明类型。
接下来,您需要声明您的闭包将要引用一些字节,并返回一个装箱的特征对象,该对象将返回一些具有相同生存期的字节,并包含具有相同生存期的引用:
struct FragMsgReceiver<'a> {
recv_dgram: &'a for<'b> FnMut(&'b mut [u8])
-> Box<MockFutureTrait<Item = &'b mut [u8]> + 'b>,
}
请参阅为什么需要Box <Iterator <Item =&Foo> +'a>? 有关+ 'a
语法的更多详细信息。
然后更新constrain_handler
以匹配:
struct FragMsgReceiver<'a> {
recv_dgram: &'a for<'b> FnMut(&'b mut [u8])
-> Box<MockFutureTrait<Item = &'b mut [u8]> + 'b>,
}
fn constrain_handler<F>(f: F) -> F
where
F: for<'b> FnMut(&'b mut [u8])
-> Box<MockFutureTrait<Item = &'b mut [u8]> + 'b>,
{
f
}
fn main() {
let mut recv_dgram = constrain_handler(|buf| Box::new(MockFuture { item: buf }));
let fmr = FragMsgReceiver {
recv_dgram: &mut recv_dgram,
};
}
如果您直接采用通用闭包,则可以简化整个过程:
struct FragMsgReceiver<R>
where
R: for<'b> FnMut(&'b mut [u8])
-> Box<MockFutureTrait<Item = &'b mut [u8]> + 'b>,
{
recv_dgram: R,
}
fn main() {
let fmr = FragMsgReceiver {
recv_dgram: |buf| Box::new(MockFuture { item: buf }),
};
}
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