迭代器中的生命周期：fn next(&'a mut self)

Question

struct Person<T> {
    letter: Letter<T>
}

struct Letter<T> {
    words: [Option<T>; 5]
}

struct EditIter<'a, T: 'a> {
    letter: &'a mut Letter<T>
}
impl<'a, T: 'a> Iterator for EditIter<'a, T> {
    type Item = Mode<'a, T>;
    fn next(&mut self) -> Option<Self::Item> {
        /* ... */
        Some(Mode::Editing(self.letter))
    }
}

enum Mode<'a, T: 'a> {
    Editing(&'a Letter<T>),
    Speaking(&'a Letter<T>)
}

错误：

src/temp.rs:16:28: 16:39 error: cannot infer an appropriate lifetime for automatic coercion due to conflicting requirements [E0495]
src/temp.rs:16         Some(Mode::Editing(self.letter))
                                          ^~~~~~~~~~~
src/temp.rs:14:5: 17:6 help: consider using an explicit lifetime parameter as shown: fn next(&'a mut self) -> Option<Self::Item>
src/temp.rs:14     fn next(&mut self) -> Option<Self::Item> {
src/temp.rs:15         /* ... */
src/temp.rs:16         Some(Mode::Editing(self.letter))
src/temp.rs:17     }

Person 拥有Letter，Person 将Letter 的可变借用转移到EditIter，EditIter 用借用的Letter 创建一个Mode 并发送它。如果您删除 Iterator 特征并按照错误消息中的说明进行操作，它可以工作，但这不会实现 Iterator 特征。

如何保存逻辑并修复错误？

Answer 1

由于您的 Mode 不需要可变引用，因此从 EditIter 中删除 mut 即可：

struct Person<T> {
    letter: Letter<T>,
}

struct Letter<T> {
    words: [Option<T>; 5],
}

struct EditIter<'a, T: 'a> {
    letter: &'a Letter<T>,
}

impl<'a, T: 'a> Iterator for EditIter<'a, T> {
    type Item = Mode<'a, T>;

    fn next(&mut self) -> Option<Self::Item> {
        Some(Mode::Editing(self.letter))
    }
}

enum Mode<'a, T: 'a> {
    Editing(&'a Letter<T>),
    Speaking(&'a Letter<T>),
}