为什么这个迭代器需要一个生命周期,我该如何给它一个生命周期?

Why does this iterator require a lifetime, and how do I give it one?

我正在尝试编译以下内容:

fn read<'a>(tokens: impl Iterator<Item=&'a str>) -> impl Iterator<Item=String> {
  tokens.map(|t| t.to_owned())
}
error[E0482]: lifetime of return value does not outlive the function call
 --> src/lib.rs:1:53
  |
1 | fn read<'a>(tokens: impl Iterator<Item=&'a str>) -> impl Iterator<Item=String> {
  |                                                     ^^^^^^^^^^^^^^^^^^^^^^^^^^
  |
note: the return value is only valid for the lifetime `'a` as defined on the function body at 1:9
 --> src/lib.rs:1:9
  |
1 | fn read<'a>(tokens: impl Iterator<Item=&'a str>) -> impl Iterator<Item=String> {
  |         ^^

游乐场link:https://play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=55460a8b17ff10e30b56a9142338ec19

令我感到奇怪的是 return 值与生命周期有关;它的内容被拥有并直接向外传递。

我看到这个博客 post:https://blog.katona.me/2019/12/29/Rust-Lifetimes-and-Iterators/

但这些都不起作用:

fn read<'a>(tokens: impl Iterator<Item=&'a str>) -> impl Iterator<Item=String> + '_ {
fn read<'a>(tokens: impl Iterator<Item=&'a str>) -> impl Iterator<Item=String> + 'a {

我怀疑我在这里误解了迭代器的一些关键方面。有什么建议吗?

我通过将 + 'a 添加到 both 迭代器来解决它:

fn read<'a>(tokens: impl Iterator<Item=&'a str> + 'a) -> impl Iterator<Item=String> + 'a {
  tokens.map(|t| t.to_owned())
}