如何返回捕获参数的过滤迭代器？答案

【问题标题】：How can I return a filtering iterator that captures an argument?如何返回捕获参数的过滤迭代器？
【发布时间】：2020-06-03 21:55:23
【问题描述】：

我正在尝试创建一个基于参数过滤切片的迭代器。

fn dates_from_iterator_ref<'a>(
    from: &'a NaiveDate,
    dates: &'a [NaiveDate],
) -> impl Iterator<Item = &'a NaiveDate> {
    dates.iter().filter(|&date| date >= from)
}

fn dates_from_iterator_val<'a>(
    from: NaiveDate,
    dates: &'a [NaiveDate],
) -> impl Iterator<Item = &'a NaiveDate> {
    dates.iter().filter(|&&date| date >= from)
}

无论我将参数作为引用还是值传递，我都会得到相同的错误：

   |
89 | fn dates_from_iterator_ref<'a>(
   |                            -- lifetime `'a` defined here
...
92 | ) -> impl Iterator<Item = &'a NaiveDate> {
   |      ----------------------------------- opaque type requires that `from` is borrowed for `'a`
93 |     dates.iter().filter(|&date| date >= from)
   |                         -------         ^^^^ borrowed value does not live long enough
   |                         |
   |                         value captured here
94 | }
   | - `from` dropped here while still borrowed

我怎样才能让它工作？

【问题讨论】：

标签： rust iterator closures

【解决方案1】：

对于值版本，您只需要通过添加 move 关键字来确保闭包通过将值移动到捕获中而不是借用它来捕获date 的值。

另外，正如@LukasKalbertodt 所指出的，你should change 从|&&date| date >= from 到|&date| date >= &from 的闭包。

pub fn dates_from_iterator_val<'a>(
    from: NaiveDate,
    dates: &'a [NaiveDate],
) -> impl Iterator<Item = &'a NaiveDate> {
    dates.iter().filter(move |&date| date >= &from)
}

对于 ref 版本，同样适用，但移动的是引用而不是值：

pub fn dates_from_iterator_ref<'a>(
    from: &'a NaiveDate,
    dates: &'a [NaiveDate],
) -> impl Iterator<Item = &'a NaiveDate> {
    dates.iter().filter(move |&date| date >= from)
}

【讨论】：