Typescript never type condition

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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”.

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