Hi,
Simple question:
Can somebody explain to me why test resolves to no?
Shouldn't it be no | yes instead?
type test = 'k' | [1] extends unknown[] ? 'yes' : 'no';
// ^? type test = "no"
#Distributivity & union of results
spark bay
next sapphire
you can read extends as "is assignable to". in that light, it's 'no' for the same reason that this is a type error:
edgy basinBOT
const a: 'k' | [1] = 'k'
const b: unknown[] = a
// ^
// Type 'string' is not assignable to type 'unknown[]'.
in case operator precedence was the source of your confusion, what you wrote is equivalent to this:
edgy basinBOT
type test = ('k' | [1]) extends unknown[] ? 'yes' : 'no';
// ^? - type test = "no"
not this:
edgy basinBOT
type test = 'k' | ([1] extends unknown[] ? 'yes' : 'no');
// ^? - type test = "k" | "yes"
I really don't get it. Why does distributivity fail here?
next sapphire
can you rephrase the question? distributivity of what, exactly?
were you expecting 'yes' | 'no' because each member of the union is checked individually?
if so, the answer is that conditional types are only distributive when the LHS is a bare type parameter
for example:
edgy basinBOT
type test2<T> = T extends unknown[] ? 'yes' : 'no';
type test3 = test2<'a' | [1]>
// ^? - type test3 = "no" | "yes"
right exactly - that's what I was looking for... carry on plz...
so in other words distributivity fails unless I use generics, right?
next sapphire
i wouldn't say it "fails". it's only designed to work when operating on a type parameter
you can use infer to create a type parameter too. i'll show an example in a moment
edgy basinBOT
type test4 = 'a' | [1] extends infer T ? T extends unknown[] ? 'yes' : 'no' : never
// ^? - type test4 = "no" | "yes"
That's perfect thank you! Exactly the information I was looking for 🙏