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,
}
}
}
我收到这个非常详细的错误消息,但我不知道如何修复代码:
error[E0495]: cannot infer an appropriate lifetime for borrow expression due to conflicting requirements
--> srcmain.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...
--> srcmain.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
--> srcmain.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...
--> srcmain.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)
--> srcmain.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
一生不是简单地'a
?
的值的类型为 Option<&'a str>
,但'a
不会使MyIterator<'a>
保持活动状态,因此它可能超出范围,并且包含的s2: String
。 所以'a
根本无法s2
活着。 (它只会让s1
活着,如果你写了fn new(s1: &'a str) -> MyIterator<'a>
,会更容易看出来)
此外,Iterator
特征的设计方式是,您永远无法在next
函数中返回对存储在Iterator
本身中的内容的引用。
相反,您可以创建一个存储值的类型,并为对该值的引用实现IntoIterator
(使用包含对存储对象的引用的单独迭代器类型)。