Dfyjkt

While working on better understanding typescript conditional typing I did hit a puzzle: i've got it working for specified type but not conditional type.

I known what i want to do can be achieved with overloading, but the new typescript tool infer and conditional typings allow us to be more generic in our type definitions and therefor not to repeat ourself that much. The following examples, cause simplified can be better expressed as overloads, but when it got more complicate it might not be the case.

Any how, on the principles i do not understand the error i made; here are the two examples:

First the types used

type In<T> = T | T
type Out<Type, T> = Type extends T ? T : T

Then for the specified (working) example:

// No generic version 'string' is passed directly
const foo = <Type extends In<string>>(
 x: Type,
): Out<Type, string> => x as any

const a = foo ('abc') // string as expected
const b = foo (['abc', 'def']) // string as expected

And finally the generic (non working) example

// Generic version type should? be inferred
const bar = <T, Type extends In<T>>(
 x: Type,
): Out<Type, T> => x as any


const c = bar ('abc') // -> Error expected string
const d = bar (['abc', 'def']) // -> Error expected string

Might be trivial but i can't find what i'm doing wrong :/

Thanks in advance
Seb

Maybe I'm misunderstanding the question (and please clarify if I do, so I can still help you), but wouldn't this simply work?

function bar<T extends number>(x: T): number
function bar<T extends string>(x: T): string
function bar<T extends any>(x: T): T
function bar<T>(x: T): T 
 return null;


const c = bar('abc') //string
const d = bar(['abc', 'def']) //string