rust - 将可变引用移出可变对象

Question

我在 enum 上有一个迭代器，它的一个变体中有一个可变引用。现在我想将此引用移出 self 并将其返回。为避免同时对同一对象有两个可变引用，我想将 self 的枚举变体更改为没有引用的变体。下面是一个代码示例:

enum Test<'a> {
    A(&'a mut usize),
    B,
}

impl<'a> Iterator for Test<'a> {
    type Item = &'a mut usize;

    fn next(&mut self) -> Option<Self::Item> {
        match *self {
            Test::A(r) => Some(r), // TODO: return mutable reference and change self to B
            Test::B => None,
        }
    }
}

fn main() {
    let mut v = 1;
    let mut it = Test::A(&mut v);
    it.next();
}

问题与Change selector in match when selector is a mutable reference有关，但该解决方案在这里不起作用。

Answer 1

我会使用 swap :

fn next(&mut self) -> Option<Self::Item> {
    if let Test::A(_) = *self {
        let mut to_swap = Test::B;

        std::mem::swap(self, &mut to_swap);
        match to_swap {
            Test::A(r)  => Some(r),
            _           => unreachable!(), // never reached
        }
    } else {
        None
    }
}

您可以使用如下辅助函数:

impl<'a> Iterator for Test<'a> {
    type Item = &'a mut usize;

    fn next(&mut self) -> Option<Self::Item> {
        if let Test::A(_) = *self {
            let mut to_swap = Test::B;

            std::mem::swap(self, &mut to_swap);
            to_swap.consume_as_a()
        } else {
            None
        }
    }
}

impl<'a> Test<'a> {
    fn consume_as_a(self) -> Option<&'a mut usize> {
        match self {
            Test::A(r)  => Some(r),
            _           => None,
        }
    }
}