CODEWITHSUNDEEP
typescript
nrofis
nrofis

Reputation: 9796

Typescript never type condition

I want to create a type with a condition based on never but I the results are unexpected (Typescript version 4.1.3).

type TypeCond<T, U> = T extends never ? {a: U} : {b: U};

let test: TypeCond<never, number>;

I have the TypeCond that has a simple condition if T extends never.

I would expect that the type of the test variable will be {a: number} but actually, the type is never.

I'm not sure why... If I replace the never with undefined or null it works as expected. But I need the condition for the never type.

Any idea why test gets a never type instead of {a: number}?

Upvotes: 12

Views: 5968

Answers (1)

kaya3
kaya3

Reputation: 51162

I believe this is a consequence of the distributive property of conditional types in Typescript. Essentially, (S | T) extends A ? B : C is equivalent to (S extends A ? B : C) | (T extends A ? B : C), i.e. the conditional type distributes over the union in the extends clause. Since every type T is equivalent to T | never, it follows that

  • T extends A ? B : C is equivalent to (T | never) extends A ? B : C,
  • Which in turn is equivalent to (T extends A ? B : C) | (never extends A ? B : C),
  • This is the same as the original type unioned with never extends A ? B : C.

So a conditional type of the form never extends A ? B : C must evaluate to never, otherwise the distributive property would be violated.

To make your type work as intended, you can use this trick:

type TypeCond<T, U> = [T] extends [never] ? {a: U} : {b: U};

This avoids the conditional type distributing over T, since [T] is not a "naked type parameter".

Upvotes: 30

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