Conditional types don't work with passed never type

[falsandtru]

cc @ahejlsberg

TypeScript Version: 2.7.0-dev.20180330

Search Terms:

Code

type C<T, U> = T extends U ? true : false;
type A = never extends void ? true : false;
type B = C<never, void>;

Expected behavior:

A and B are true.

Actual behavior:

A is true, B is never.

Playground Link:

Related Issues:

[jack-williams]

I believe this is working as intended: C is distributive but A is not. C is being distributed zero times as never is empty.

[falsandtru]

A is inlined B. Inlining must not make any different behavior.

[mhegazy]

A is inlined B. Inlining must not make any different behavior.

not really. a type is distributed if it has a naked type parameter in the condition close. so

type inline = number | string extends string ? true : false; // false

type deferred<T> = T extends string ? true : false;
type d = deferred<number | string>;  // boolean

The rational here that using a type parameter in the condition position is like a function on the input, and is expected to distribute, where as an actual union type is mean to be used as is.

[falsandtru]

I understood, this is not inlining. A passed union type is not determined as its exact union type yet. It also can be a subtype.

[falsandtru]

A workaround:

type C<T, U> = [T] extends [never] ? 1 : T extends U ? 1 : 0;
type a = C<never, void>; // 1
type b = C<never, never>; // 1
type c = C<boolean, true>; // 0 | 1