`for` borrowing lasts after the end of the for loop

[progval]

Hi,

The following code:

fn method(vector: &mut Vec<u64>) -> &u64 {
    {
        for item in vector.iter() {
            return item
        }
    }

    vector.last_mut().unwrap()
}

Raises an error about vector.iter()'s borrow conflicting with vector.last_mut():

error[E0502]: cannot borrow `*vector` as mutable because it is also borrowed as immutable
 --> <anon>:8:5
  |
3 |         for item in vector.iter() {
  |                     ------ immutable borrow occurs here
...
8 |     vector.last_mut().unwrap()
  |     ^^^^^^ mutable borrow occurs here
9 | }
  | - immutable borrow ends here

However, I would have expected the first borrow to end with the first closing bracket.

Playground URL: https://play.rust-lang.org/?gist=36c206ebc1f8c96b96978a628b5c09a9&version=stable&backtrace=0

[steveklabnik]

I believe this is a duplicate, but I'm not sure.

[hanna-kruppe]

This is because of lexical lifetimes and is thus covered by rust-lang/rfcs#811. item is returned, so it must be valid for longer than the loop (longer than the function, actually) and consequently the vector is also borrowed for as long. This isn't a real problem because this only applies to the return path, but because lifetimes are lexical, the compiler only has one choice for the lifetime of that borrow (namely, the lifetime encompassing the whole function).

[steveklabnik]

I was not 100% sure, but this was also a suspicion I had. Thanks for the report, @progval , but yes, it's covered by that linked issue.

[nox]

But the whole loop is in a block by itself, why would the borrow need to survive longer than the enclosing block? Oh I see. Very weird to call this "not a real problem" though, @rkruppe.

[steveklabnik]

You're reading "not a real problem" backwards; @rkruppe is saying that your code is fine, that it doesn't have an actual problem. Borrowck can't understand that, though.

[nox]

Oh I see, thanks for the explanation!