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Generic conditional type T extends never ? 'yes' : 'no' resolves to never when T is never. #31751
Description
TypeScript Version: 3.6.0-dev.20190603
Search Terms:
never, extends, generic, conditional type
Code
type MakesSense = never extends never ? 'yes' : 'no' // Resolves to 'yes'
type ExtendsNever<T> = T extends never ? 'yes' : 'no'
type MakesSenseToo = ExtendsNever<{}> // Resolves to 'no'
type Huh = ExtendsNever<never> // Expect to resolve to 'yes', actually resolves to never
Expected behavior:
Huh should resolve to 'yes', as MakesSense does.
Actual behavior:
Huh resolves to never.
For another variant, you can change ExtendsNever to
type ExtendsNever<T extends never> = T extends never ? 'yes' : 'no'
and ExtendsNever still resolves to never.
Playground Link:
Here
Related Issues:
None
Activity
dragomirtitian commentedon Jun 4, 2019
This is the expected behavior,
ExtendsNeveris a distributive conditional type. Conditional types distribute over unions. Basically ifTis a unionExtendsNeveris applied to each member of the union and the result is the union of all applications (ExtendsNever<'a' | 'b'> == ExtendsNever<'a' > | ExtendsNever<'b'>).neveris the empty union (ie a union with no members). This is hinted at by its behavior in a union'a' | never == 'a'. So when distributing overnever,ExtendsNeveris never applied, since there are no members in this union and thus the result is never.If you disable distribution for
ExtendsNever, you get the behavior you expect:rickwebiii commentedon Jun 4, 2019
Thanks for the clarification. I should really get around to reading my Homotopy Type Theory book.
fatcerberus commentedon Jun 5, 2019
Why is
nevertreated as an empty union butunknownisn’t treated as a “union of everything”? For symmetry I would expectT extends any ? ...whereT = unknownto take both branches of the conditional type, the same wayT = nevertakes neither.I understand that
anyworks this way butanyisn’t really a “union of all types” in the usual sense - it also behaves as an “intersection of all types” when placed in a contravariant position so it’s kind of its own thing entirely and I expect it to act weird. 😉jack-williams commentedon Aug 11, 2019
Bruce Pascoe (@fatcerberus); Discussed in #27418
Alexsey commentedon Jul 27, 2021
I've got here because I was unable to understand why
MakesSenseandHuhare getting evaluated differently i.e. why both of them are notnever. So probably here is a good place to post the answer:According to the handbook chapter on Distributed Conditional Types
It means that when
Huhis evaluated, thenevervalue ofTis getting distributed and things are happening as Titian Cernicova-Dragomir (@dragomirtitian) explained. But evaluation ofMakesSensedoesn't include generics, so Conditional Type doesn't get distributed and so it acts as "expected" -neveris getting compared withneverZhanghaoH commentedon Jul 19, 2022
but type-challenges-01097 ，its last test example thinks Never is not Union Type
Alexsey commentedon Jul 19, 2022
ZhanghaoH you can think of the type challenge problem not as "isUnion" but as "isDistributiveUnion" or "isUnionOfMoreThenOneElement".
1is also a union - a union of 1 element. Andneveris a union - an empty union, or a union of 0 elementsUnions of 0 elements, 1 element, and 1+ elements behave differently in TS, and the problem is about detecting the third, non-singular case
ZhanghaoH commentedon Jul 20, 2022
I see...thanks for your reply !