Inferring ReturnType<T> from function with arguments falling back to any

[dmtrKovalenko]

TypeScript Version: 3.3.0-dev.20190119

Search Terms:
Infering type from function return type
Infering from return type
Argument breaks return type inferring

Code

export const createStyles = <
  K extends ReturnType<T>,  
  T extends (theme: { type: string }) => StyleSheet.NamedStyles<K> | StyleSheet.NamedStyles<any>,
>(
  fn: T
) => fn;

const styles = createStyles(() => ({
  logo: {
    width: 70,
    height: 70,
  },
}));

// here logo type is properly inferred from object
styles().logo.width


const styles = createStyles((theme) => ({
  logo: {
    width: 70,
    height: 70,
  },
}));

// here styles function signature is (theme: { type: string })  =>  any
styles({ type: 'dark' }).logo.width

Expected behavior:
It is possible to infer the return type of the function for both situations

Actual behavior:
Type inferring only for the function that doesn't have any parameters and falling to any if any parameters applied

Playground Link:
https://codesandbox.io/s/p3414yomnx

Related Issues:

[RyanCavanaugh]

Seems like you just have too many type parameters. This code works better:

export const createStyles = <
  T extends (
    theme: { type: string }
  ) => StyleSheet.NamedStyles<unknown>
>(
  fn: T
) => fn;

[dmtrKovalenko]

@RyanCavanaugh Thanks for clarifying. But for react native it's required to infer type through NamedStyles to be sure that strings like textAlign are proper enum values.

And here is a question - is it possible to make it works with inferring ReturnType through NamedStyles

[dmtrKovalenko]

I understand that here we getting circular dependency between types, referring each other. But maybe there is some way to not fall back to any?

[dmtrKovalenko]

And why it's breaking only if function has argument 🤔

[dmtrKovalenko]

Duplicate of #241