Conditional type doesn't think T & U extends U

[jcreedcmu]

TypeScript Version: 3.3.0-dev.20181201

Search Terms: Conditional Types, Intersection

Code

type NumTest<S> = S extends number ? string : never;
type Str<T> = NumTest<T & number>; // should be string, I would think?
function foo<T>() {
  const x: Str<T> = "hello";
}

Expected behavior:
No errors. More narrowly, I expect T & number to be understood as extending number.

Notably, if I inline the definition of NumTest<S>, i.e.
replace the above definition of Str with the following:

type Str<T> = (T & number) extends number ? string : never;

then the error below vanishes.

Actual behavior:

ex.ts:4:9 - error TS2322: Type '"hello"' is not assignable to type 'NumTest<T & number>'.

4   const x: Str<T> = "hello";
          ~


Found 1 error.

Playground Link

[jack-williams]

The conditional type

type NumTest<S> = S extends number ? string : never;`

is a distributive one, meaning that instantiating S to never will return never, not string.

For your example:

type NumTest<S> = S extends number ? string : never;
type Str<T> = NumTest<T & number>; // could be never
function foo<T>() {
  const x: Str<T> = "hello";
}

Pick T to be never, so that T & never === never, and therefore NumTest<never> === never. The string "hello" is not assignable to never, hence the error. If you disable distribution you get the expected behaviour.

type NumTest<S> = [S] extends [number] ? string : never;
type Str<T> = NumTest<T & number>;
function foo<T>(): string {
  const x: Str<T> = "hello";
  return x;
}

Note that

type Str<T> = (T & number) extends number ? string : never;

works precisely because it is not a distributive conditional type. The check type is not a naked type parameter, rather it is an intersection type.

[jcreedcmu]

Ah, I see. Thanks for the clear explanation. I indeed forgot to consider the never case of distributivity.