函数返回一个闭包不在我的过滤器内工作

Question

我不能在不使用闭包的情况下编译它.我试图让函数apply首先返回正确的闭包.

#![feature(conservative_impl_trait)]
#![allow(dead_code)]

fn accumulate<'a>(tuples: &[(&'a str, &Fn(i32) -> bool)], i: i32) {

    // this works
    let _ = tuples.iter().filter(|t| apply(second, i)(t));

    // this doesn't
    //let f = apply(second, i);
    //let _ = tuples.iter().filter(f);

    //this works as well

    let f  = |t: &&(_,_)| apply(second, i)(t);
    let _ = tuples.iter().filter(f);
}

fn apply<A, B, C, F, G>(mut f: F, a: A) -> impl FnMut(B) -> C
         where F: FnMut(B) -> G,
               G: FnMut(A) -> C,
               A: Clone
{
    move |b| f(b)(a.clone())
}


fn second<A, B: ?Sized>(&(_, ref second): &(A, B)) -> &B {
    second
}

fn main()  {}

我能做些什么来做我喜欢的apply工作？

Answer 1

首先,让我说这个问题与impl Trait语法的使用无关.我将闭包转换为命名结构并获得相同的结果.

那么,让我们来看看你想要做的代码:

let f = apply(second, i);
let _ = tuples.iter().filter(f);

编译器有什么说法呢？

error[E0277]: the trait bound `for<'r> impl std::ops::FnMut<(&(_, _),)>: std::ops::FnMut<(&'r &(&str, &std::ops::Fn(i32) -> bool),)>` is not satisfied
  --> <anon>:11:27
   |
11 |     let _ = tuples.iter().filter(f);
   |                           ^^^^^^ trait `for<'r> impl std::ops::FnMut<(&(_, _),)>: std::ops::FnMut<(&'r &(&str, &std::ops::Fn(i32) -> bool),)>` not satisfied

error[E0277]: the trait bound `for<'r> impl std::ops::FnMut<(&(_, _),)>: std::ops::FnOnce<(&'r &(&str, &std::ops::Fn(i32) -> bool),)>` is not satisfied
  --> <anon>:11:27
   |
11 |     let _ = tuples.iter().filter(f);
   |                           ^^^^^^ trait `for<'r> impl std::ops::FnMut<(&(_, _),)>: std::ops::FnOnce<(&'r &(&str, &std::ops::Fn(i32) -> bool),)>` not satisfied

好的,所以我们有X型,它需要实现特性Y,但事实并非如此.但让我们仔细看看:

for<'r> impl
std::ops::FnMut<(&(_, _),)>:
std::ops::FnMut<(&'r &(_, _),)>

啊哈!filter期望一个接受对元组引用的引用的函数,而我们传入的函数接受对元组的引用.filter传递对引用的引用,因为tuples.iter()迭代引用,并filter传递对引用的引用.

好吧,让我们改变second接受引用引用的定义:

fn second<'a, A, B: ?Sized>(&&(_, ref second): &&'a (A, B)) -> &'a B {
    second
}

编译器仍然不满意:

error[E0277]: the trait bound `for<'r> impl std::ops::FnMut<(&&(_, _),)>: std::ops::FnMut<(&'r &(&str, &std::ops::Fn(i32) -> bool),)>` is not satisfied
  --> <anon>:11:27
   |
11 |     let _ = tuples.iter().filter(f);
   |                           ^^^^^^ trait `for<'r> impl std::ops::FnMut<(&&(_, _),)>: std::ops::FnMut<(&'r &(&str, &std::ops::Fn(i32) -> bool),)>` not satisfied

error[E0271]: type mismatch resolving `for<'r> <impl std::ops::FnMut<(&&(_, _),)> as std::ops::FnOnce<(&'r &(&str, &std::ops::Fn(i32) -> bool),)>>::Output == bool`
  --> <anon>:11:27
   |
11 |     let _ = tuples.iter().filter(f);
   |                           ^^^^^^ expected bound lifetime parameter , found concrete lifetime
   |
   = note: concrete lifetime that was found is lifetime '_#24r

expected bound lifetime parameter , found concrete lifetime... 那是什么意思？

f的类型是一种实现的类型FnMut(&'c &'b (&'a str, &Fn(i32) -> bool)) -> bool.在致电apply,B == &'c &'b (&'a str, &Fn(i32) -> bool)和C == bool.注意这里B是一个固定类型 ; 'c代表一个固定的寿命,称为具体的寿命.

我们来看看filter签名:

fn filter<P>(self, predicate: P) -> Filter<Self, P> where
    Self: Sized, P: FnMut(&Self::Item) -> bool,

在这里,P必须实施FnMut(&Self::Item) -> bool.实际上,这种语法是简写for<'r> FnMut(&'r Self::Item) -> bool.这里.'r是一个绑定生命周期参数.

所以,问题是我们实现的FnMut(&'c &'b (&'a str, &Fn(i32) -> bool)) -> bool函数没有实现for<'r> FnMut(&'r Self::Item) -> bool.我们需要一个实现的功能for<'c> FnMut(&'c &'b (&'a str, &Fn(i32) -> bool)) -> bool.到目前为止,唯一的方法是这样写apply:

fn apply<A, B, C, F, G>(mut f: F, a: A) -> impl FnMut(&B) -> C
         where F: FnMut(&B) -> G,
               G: FnMut(A) -> C,
               A: Clone
{
    move |b| f(b)(a.clone())
}

或更明确的版本:

fn apply<A, B, C, F, G>(mut f: F, a: A) -> impl for<'r> FnMut(&'r B) -> C
         where F: for<'r> FnMut(&'r B) -> G,
               G: FnMut(A) -> C,
               A: Clone
{
    move |b| f(b)(a.clone())
}

如果Rust最终支持更高级的类型,则可能有更优雅的方法来解决此问题.