What does a match on an empty enum return?

Question

When reading Rust's convert.rs, I encountered the following code:

#[unstable(feature = "try_from", issue = "33417")]
#[derive(Clone, Copy, Debug, Eq, Hash, Ord, PartialEq, PartialOrd)]
pub enum Infallible {}

#[unstable(feature = "try_from", issue = "33417")]
impl fmt::Display for Infallible {
    fn fmt(&self, _: &mut fmt::Formatter) -> fmt::Result {
        match *self {
        }
    }
}

Infallible is an empty enum with no variants. What does match *self {} return?

score 11 · Answer 1 · answered Dec 01 '17 at 08:59

11

Since Infallible has no possible values, you can never have an instance of it. This means matching on it can never happen. Rust represents this by making matching on an empty enum yield the ! type, which is a builtin type that has no values.

This type coerces to any other type, because the statement can never be reached, because you'd need a value of type Infallible for that, which you can't have out of obvious reasons.

answered Dec 01 '17 at 08:59

oli_obk

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If it's unreachable, why `match *self {}` is used here instead of `unreachable!()`? – Cheng-Chang Wu Dec 01 '17 at 09:23
@Cheng-ChangWu and why would it implement `Display` anyway? – ljedrz Dec 01 '17 at 09:39
1

@Cheng-ChangWu that's probably because `match *self {}` produces no code, while `unreachable!()` produces panic code that will need to get optimized out – oli_obk Dec 01 '17 at 09:42

What does a match on an empty enum return?

1 Answers1