一个关于迭代器和所有权的问题？

我是 Rust 的新手，跟着 Rust 权威指南和视频学到了并发章节，有个例子遇到了点问题：

use std::{sync::mpsc, thread, time::Duration};

fn main() {
  let (tx, rx) = mpsc::channel();

  thread::spawn(move || {
    let v = vec![
      String::from("Hello"),
      String::from("I'm XiaoMing"),
      String::from("I come from China"),
      String::from("Nice to meet you"),
    ];

    for s in v.iter() {
      tx.send(s).unwrap();
      thread::sleep(Duration::from_millis(300));
    }
  });

  for msg in rx {
    println!("{}", msg);
  }
}

我的 VSCode 在 v.iter() 处报了错：

`v` does not live long enough
borrowed value does not live long enoughrustcE0597
main.rs(18, 3): `v` dropped here while still borrowed
main.rs(4, 8): lifetime `'1` appears in the type of `tx`
main.rs(15, 7): argument requires that `v` is borrowed for `'1`

我知道 for in 循环会取得迭代器的所有权，iter 方法的第一个参数是 &self，它没有取得 v 的所有权，迭代器的所有权转移了，为什么 v 会失效？

另外如果我不调用 iter 直接用 v 去循环就能通过，这两种有什么区别吗？

====== 补充下在 rust playground 里的错误信息，可能有帮助：

   Compiling playground v0.0.1 (/playground)
error[E0597]: `v` does not live long enough
  --> src/main.rs:14:14
   |
4  |   let (tx, rx) = mpsc::channel();
   |        -- lifetime `'1` appears in the type of `tx`
...
14 |     for s in v.iter() {
   |              ^^^^^^^^ borrowed value does not live long enough
15 |       tx.send(s).unwrap();
   |       ---------- argument requires that `v` is borrowed for `'1`
...
18 |   });
   |   - `v` dropped here while still borrowed

For more information about this error, try `rustc --explain E0597`.
error: could not compile `playground` due to previous error