Primitive type `boolean` does not work with conditional types

[benkeil]

Bug Report

🔎 Search Terms

conditional type extends boolean

🕗 Version & Regression Information

4.3.2

⏯ Playground Link

Playground link with relevant code

💻 Code

type TestFunction<I, O> = I extends void ? () => O : (result: I) => O;

const numberFunction: TestFunction<number, void> = () => {};
numberFunction(1);
const booleanFunction1: TestFunction<boolean, void> = () => {};
booleanFunction1(true);
const booleanFunction2: TestFunction<Boolean, void> = () => {};
booleanFunction2(true);

🙁 Actual behavior

Looks like the keyword extends does not work with primitive type boolean but with Class Boolean and also with primitive type number.

TS2345: Argument of type 'boolean' is not assignable to parameter of type 'never'.

🙂 Expected behavior

booleanFunction2 should act like booleanFunction1

[MartinJohns]

boolean do work with conditional types, but they (intentional) are distributive.

[anuraghazra]

Right, we can avoid distribution of types by wrapping it in a tuple.

related to #29368 (comment)

type TestFunction<I, O> = [I] extends [void] ? () => O : (result: I) => O;

const numberFunction: TestFunction<number, void> = () => {};
numberFunction(1);
const booleanFunction1: TestFunction<boolean, void> = () => {};
booleanFunction1(true);
const booleanFunction2: TestFunction<Boolean, void> = () => {};
booleanFunction2(true);

[MartinJohns]

Right, we can avoid distribution of types by wrapping it in a tuple.

As shown in the linked documentation.

[benkeil]

First of all thank you for pointing out. But can you please explain why boolean acts different and what exactly happens here?

[fatcerberus]

boolean acts differently because, in practice, it’s treated as a union of true | false. When a conditional type is applied to a type union (specifically only when the input to the conditional type is a bare generic type parameter), it “distributes” over the members of the union—think Array#map().