如何分配给匹配分支内的匹配表达式中使用的变量？

Question

我正在尝试实现一个通用函数 join()，它可以在迭代器的任何迭代器上工作。 next() 方法实现中的 match 表达式中的借用检查器有问题。这是我的代码的简化版本：

pub struct Join<I>
where
    I: Iterator,
    I::Item: IntoIterator,
{
    outer_iter: I,
    inner_iter: Option<<I::Item as IntoIterator>::IntoIter>,
}

impl<I> Join<I>
where
    I: Iterator,
    I::Item: IntoIterator,
{
    pub fn new(mut iter: I) -> Join<I> {
        let inner_iter = iter.next().map(|it| it.into_iter());
        Join {
            outer_iter: iter,
            inner_iter,
        }
    }
}

impl<I> Iterator for Join<I>
where
    I: Iterator,
    I::Item: IntoIterator,
{
    type Item = <I::Item as IntoIterator>::Item;

    fn next(&mut self) -> Option<Self::Item> {
        loop {
            match &mut self.inner_iter {
                Some(ref mut it) => match it.next() {
                    Some(x) => {
                        return Some(x);
                    }
                    None => {
                        self.inner_iter = self.outer_iter.next().map(|it| it.into_iter());
                    }
                },
                None => {
                    return None;
                }
            }
        }
    }
}

pub trait MyItertools: Iterator {
    fn join(self) -> Join<Self>
    where
        Self: Sized,
        Self::Item: IntoIterator,
    {
        Join::new(self)
    }
}

impl<I> MyItertools for I where I: Iterator {}

#[cfg(test)]
mod test {
    use super::MyItertools;

    #[test]
    fn it_works() {
        let input = [[1], [2]];
        let expected = [&1, &2];
        assert_eq!(input.iter().join().collect::<Vec<_>>(), expected);
    }
}

错误文本：

error[E0506]: cannot assign to `self.inner_iter` because it is borrowed
  --> src/main.rs:39:25
   |
33 |             match &mut self.inner_iter {
   |                        --------------- borrow of `self.inner_iter` occurs here
...
39 |                         self.inner_iter = self.outer_iter.next().map(|it| it.into_iter());
   |                         ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ assignment to borrowed `self.inner_iter` occurs here

我明白为什么借用检查器会抱怨我的代码，但我没有找到好的解决方案，只有一个丑陋的解决方法：

fn next(&mut self) -> Option<Self::Item> {
    loop {
        match self.inner_iter.take() {
            Some(mut it) => {
                match it.next() {
                    Some(x) => { self.inner_iter = Some(it); return Some(x); }
                    None => { self.inner_iter = self.outer_iter.next().map(|it| it.into_iter()); }
                }
            }
            None => { return None; }
        }
    }
}

我想像这样的情况经常发生；我如何重写我的代码来处理或避免它们？

Answer 1

在这种情况下，我发现将代码分两部分编写很有用：首先收集数据，然后更新可变变量：

fn next(&mut self) -> Option<Self::Item> {
    loop {
        //collect the change into a local variable
        let ii = match &mut self.inner_iter {
            Some(ref mut it) => {
                match it.next() {
                    Some(x) => { return Some(x); }
                    None => self.outer_iter.next().map(|it| it.into_iter())
                }
            }
            None => { return None; }
        };
        //self.inner_iter is no longer borrowed, update
        self.inner_iter = ii;
    }
}

事实上，所有不修改 inner_iter 的分支都执行 return 使得代码更简单。

Answer 2

这里是问题的简单重现：

fn main() {
    let mut a = (42, true);
    match a {
        (ref _i, true) => a = (99, false),
        (ref _i, false) => a = (42, true),
    }
    println!("{:?}", a);
}

error[E0506]: cannot assign to `a` because it is borrowed
 --> src/main.rs:4:27
  |
4 |         (ref _i, true) => a = (99, false),
  |          ------           ^^^^^^^^^^^^^^^ assignment to borrowed `a` occurs here
  |          |
  |          borrow of `a` occurs here

error[E0506]: cannot assign to `a` because it is borrowed
 --> src/main.rs:5:28
  |
5 |         (ref _i, false) => a = (42, true),
  |          ------            ^^^^^^^^^^^^^^ assignment to borrowed `a` occurs here
  |          |
  |          borrow of `a` occurs here

这是基于 AST 的借用检查器的弱点。当non-lexical lifetimes are enabled, this works as-is。增强的基于 MIR 的借用检查器可以看到在您尝试替换匹配的变量时没有借用它。

就其价值而言，你的 join 只是一个 flat_map:

input.iter().flat_map(|x| x)

或者 flatten:

input.iter().flatten()

你可以看看这些implement next换个思路：

fn next(&mut self) -> Option<Self::Item> {
    loop {
        if let Some(v) = self.inner_iter.as_mut().and_then(|i| i.next()) {
            return Some(v);
        }
        match self.outer_iter.next() {
            Some(x) => self.inner_iter = Some(x.into_iter()),
            None => return None,
        }
    }
}

这清楚地表明迭代器值并非真正借自inner_iter。

如果不看 flatten，我会选择明确指出 taking 和 Option 没有重叠借用，如果是 Some 则恢复它，正如你所做的那样：

match self.inner_iter.take() {
    Some(mut it) => match it.next() {
        Some(x) => {
            self.inner_iter = Some(it);
            return Some(x);
        }
        None => {
            self.inner_iter = self.outer_iter.next().map(|it| it.into_iter());
        }
    },
    None => {
        return None;
    }
}

如何分配给匹配分支内的匹配表达式中使用的变量？

How to assign to the variable used in match expression inside a match branch?

match

lifetime

rust

borrow-checker