Typescript simple generic inference issue

Question

What's wrong with this ?
Shouldn't be a string ?

const valueGetter = <T extends {value:V}, V>(o:T) => ():V => o.value 
const myValueGetter = valueGetter({x:1, value:'a string'})  // const myValueGetter: () => {}
const be = myValueGetter() // const be: {}

try it live

What makes you think it's an object? console.log(typeof be); — smac89
– smac89, Commented Jun 21, 2018 at 20:56

artem · Accepted Answer · 2018-06-21 21:08:59Z

1

I don't know why, but somehow TypeScript does not recognize the dependency between V and the type of value in T.

In such cases, it helps to be explicit that the desired type comes from a type of a member in T:

const valueGetter = <T extends {value: V}, V>(o: T) => ((): T['value'] => o.value) 
const myValueGetter = valueGetter({x:1, value:'a string'})  // const myValueGetter: () => string
const be = myValueGetter() // const be: string

answered Jun 21, 2018 at 21:08

artem

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3 Comments

Titian Cernicova-Dragomir Over a year ago

You don't really need the type parameter V anymore, you can just use T extends {value: any}.

aleclofabbro Over a year ago

Nice how-to make it work, and @TitianCernicova-Dragomir you're right! despite this doesn't explain how that bizarre inference occurred...

Titian Cernicova-Dragomir Over a year ago

@aleclofabbro not that bizarre, Typescript doesn't really do dependant generic type parameters, it tries to get away with the least restrictive type posibile when presented with casses such as yours. Conditional types and type queries are the way to work around this.

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