rust - 尝试实现借入一些数据并拥有其他数据的迭代器时发生编译错误

This question already has answers here:

Can I write an Iterator that mutates itself and then yields a reference into itself?

(1个答案)



How do I write an iterator that returns references to itself?

(3个答案)


3年前关闭。




我正在尝试实现一个迭代器:

struct MyIterator<'a> {
    s1: &'a str,
    s2: String,

    idx: usize,
}

impl<'a> MyIterator<'a> {
    fn new(s1: &str) -> MyIterator {
        MyIterator {
            s1: s1,
            s2: "Rust".to_string(),

            idx: 0,
        }
    }
}

impl<'a> Iterator for MyIterator<'a> {
    type Item = &'a str;

    fn next(&mut self) -> Option<Self::Item> {
        self.idx += 1;

        match self.idx {
            1 => Some(self.s1),
            2 => Some(&self.s2),
            _ => None,
        }
    }
}

而且我收到了非常详细的错误消息，但是我不知道如何修复代码:

error[E0495]: cannot infer an appropriate lifetime for borrow expression due to conflicting requirements
  --> src\main.rs:39:23
   |
39 |             2 => Some(&self.s2),
   |                       ^^^^^^^^
   |
note: first, the lifetime cannot outlive the anonymous lifetime #1 defined on the method body at 34:5...
  --> src\main.rs:34:5
   |
34 | /     fn next(&mut self) -> Option<Self::Item> {
35 | |         self.idx + 1;
36 | |
37 | |         match self.idx {
...  |
41 | |         }
42 | |     }
   | |_____^
note: ...so that reference does not outlive borrowed content
  --> src\main.rs:39:23
   |
39 |             2 => Some(&self.s2),
   |                       ^^^^^^^^
note: but, the lifetime must be valid for the lifetime 'a as defined on the impl at 31:1...
  --> src\main.rs:31:1
   |
31 | / impl<'a> Iterator for MyIterator<'a> {
32 | |     type Item = &'a str;
33 | |
34 | |     fn next(&mut self) -> Option<Self::Item> {
...  |
42 | |     }
43 | | }
   | |_^
note: ...so that types are compatible (expected std::iter::Iterator, found std::iter::Iterator)
  --> src\main.rs:34:46
   |
34 |       fn next(&mut self) -> Option<Self::Item> {
   |  ______________________________________________^
35 | |         self.idx + 1;
36 | |
37 | |         match self.idx {
...  |
41 | |         }
42 | |     }

为什么s2生存期不只是'a？

最佳答案

返回的值的类型为Option<&'a str>，但是'a不能使MyIterator<'a>保持 Activity 状态，因此它可能超出范围，并且包含了s2: String。因此，'a根本无法使s2保持 Activity 状态。 (它只能使s1保持 Activity 状态，这样可以更轻松地查看是否已编写fn new(s1: &'a str) -> MyIterator<'a>)

此外， Iterator 特质的设计方式是，您可以永不在Iterator函数中返回对next本身中存储的内容的引用。

相反，您可以创建一个存储该值的类型，并实现 IntoIterator 以对其进行引用(使用一个单独的迭代器类型，该类型包含对存储对象的引用)。

关于rust - 尝试实现借入一些数据并拥有其他数据的迭代器时发生编译错误，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/47054316/