我在尝试编译下面的Rust代码时遇到了一对奇怪的错误 . 在搜索有类似问题的其他人时,我遇到了another question with the same combination of (seemingly opposing) errors,但无法将解决方案从那里推广到我的问题 .
基本上,我似乎在Rust的所有权系统中缺少一个微妙的东西 . 在这里尝试编译(非常削减)代码:
struct Point {
x: f32,
y: f32,
}
fn fold<S, T, F>(item: &[S], accum: T, f: F) -> T
where
F: Fn(T, &S) -> T,
{
f(accum, &item[0])
}
fn test<'a>(points: &'a [Point]) -> (&'a Point, f32) {
let md = |(q, max_d): (&Point, f32), p: &'a Point| -> (&Point, f32) {
let d = p.x + p.y; // Standing in for a function call
if d > max_d {
(p, d)
} else {
(q, max_d)
}
};
fold(&points, (&Point { x: 0., y: 0. }, 0.), md)
}
我收到以下错误消息:
error[E0631]: type mismatch in closure arguments
--> src/main.rs:23:5
|
14 | let md = |(q, max_d): (&Point, f32), p: &'a Point| -> (&Point, f32) {
| ---------------------------------------------------------- found signature of `for<'r> fn((&'r Point, f32), &'a Point) -> _`
...
23 | fold(&points, (&Point { x: 0., y: 0. }, 0.), md)
| ^^^^ expected signature of `for<'r> fn((&Point, f32), &'r Point) -> _`
|
= note: required by `fold`
error[E0271]: type mismatch resolving `for<'r> <[closure@src/main.rs:14:14: 21:6] as std::ops::FnOnce<((&Point, f32), &'r Point)>>::Output == (&Point, f32)`
--> src/main.rs:23:5
|
23 | fold(&points, (&Point { x: 0., y: 0. }, 0.), md)
| ^^^^ expected bound lifetime parameter, found concrete lifetime
|
= note: required by `fold`
(A Rust Playground link for this code, for convenience.)
在我看来,我正在提供给 fold
的功能应该正确地进行类型检查......我在这里缺少什么,我该如何解决它?
1 回答
简短版本是,如果闭包内联写入或存储为变量,则推断的生命周期之间存在差异 . 写内联闭包并删除所有无关的类型:
另外,我必须从输入数组中提取
first
值 - 你可以't return a reference to a local variable. There'不需要方法的生命周期参数;他们将被推断 .要实际获取要编译的代码,您需要提供有关
fold
方法的更多信息 . 具体来说,您必须指出传递给闭包的引用与传入的参数具有相同的生命周期 . 否则,它可能只是对局部变量的引用:相关的Rust问题是#41078 .