Mutable borrows in an else if after an if let are disallowed

[ftxqxd]

fn main() {
    let mut f = Some(1);
    if let Some(a) = f {
    } else if f.as_mut().is_some() {}
}

Gives:

<anon>:4:15: 4:16 error: cannot mutably borrow in a pattern guard [E0301]
<anon>:4     } else if f.as_mut().is_some() {}
                       ^
note: in expansion of if let expansion
<anon>:3:5: 4:38 note: expansion site

This is an unexpected result of how if let desugaring works, presumably. It’s unexpected because mutable borrowing is normally allowed in else if conditions, but because if let seems to desugar a following else if into a pattern guard, this is disallowed.

To me it would make more sense for this to be treated like any other else if, and be permitted, or at the very least have a clearer message (because it’s not obvious (and is really just an implementation detail) how an else if condition is a pattern guard).

[sfackler]

if let desugars to a match, which is how that error message is showing up:

match f {
    Some(a) => {}
    _ if f.as_mut().is_some() {}
    _ => {}
}

[arielb1]

duplicate #6393

[arielb1]

actually, not.

Why is this a pattern guard rather than an if in the _ anyway?

[Aatch]

@arielb1 it's a pattern guard because this works:

if let Some(1) = foo {

} else if check(&foo) {

} else if let Some(a) = foo {

}

Obviously this doesn't require desugaring to a single match, but it does explain why it does right now.