为什么不直接作为函数参数调用时，此迭代器会掉线？

Question

在以下示例中，我对Rust的行为感到困惑。

此示例失败：

let mut coll = vec![1, 2, 3].iter();
some_consuming_fn(&mut coll);

error[E0716]: temporary value dropped while borrowed
 --> src/main.rs:4:20
  |
4 |     let mut coll = vec![1, 2, 3].iter();
  |                    ^^^^^^^^^^^^^       - temporary value is freed at the end of this statement
  |                    |
  |                    creates a temporary which is freed while still in use
5 |     some_consuming_fn(&mut coll);
  |                       --------- borrow later used here
  |
  = note: consider using a `let` binding to create a longer lived value
  = note: this error originates in a macro outside of the current crate (in Nightly builds, run with -Z external-macro-backtrace for more info)

但是，如果将可迭代转换作为函数参数调用，则一切正常。

let coll = vec![1, 2, 3];
some_consuming_fn(&mut coll.iter());

我对已变成可迭代的集合的所有权规则感到困惑。为什么对some_iterable.iter()的呼叫在被呼叫后立即中断？

供参考，some_consuming_fn具有签名：

some_consuming_fn(x: &mut Iter<u32>) -> Foo;