alt Hacker NewsWhy is it bad? Implicit integer conversions are generally bad because they can produce unexpected behavior at runtime and obstruct what’s really happening, but that doesn’t seem to be what’s happening here.
Never is a standard type in many languages and is at the bottom of the type hierarchy because it’s a subtype of every type. Never isn’t implicitly converted any more than `&’a A` is “implicitly converted” into a `&’b B`, where `’a` subsumes `’b`. There’s no runtime conversion because there will never be an instance of never—it represents the value of a computation that never completes by definition.
I think what you mean to say is that implicit runtime conversions are bad, not that all subtyping is bad.
Replies
No I’m not talking about runtime conversions. I’m talking about conversions that happen at type inference time.
Rust is not a subtyping based language, except for traits and lifetimes. So statements like never being at the bottom of the type hierarchy is irrelevant here even though it is correct. If Rust had higher rank types the never type is also (forall a. a) but still it doesn’t matter. It is simply surprising for a type to be converted implicitly according to subtyping rules other than for traits and lifetimes.
You’re using “subtype” in two distinct, but related, senses here, and I think this should be clarified.
From a more category-theoretic perspective, a type A is a “subtype” of a type B when there is an embedding of A inside B. In this sense, `!` is a subtype of every type (which is its universal property). But this definition also grants you that `String` is a subtype of `BigInt`, because strings can be coded as bit sequences which can be coded in `BigInt`, which may or may not be what you expect.
From a programming languages perspective – and this is the terminology generally used in Rust – a type A is a “subtype” of a type B when `a: A` implies that `a: B`. In this sense, `!` is only a subtype of itself; although it coerces to any other type, it’s not _literally_ of that type, the coercion is just invisible in syntax. Importantly, if A is a subtype of B then `Vec<A>` is a subtype of `Vec<B>` – but `Vec<!>` is definitely not a subtype of `Vec<T>`, since they may have totally different layouts in memory (the former not allocating at all, while the latter potentially allocating).