Function type argument inference indirectly lost with generic function

[andykais]

TypeScript Version: 3.4.0-dev.20190213 (and 3.3.3)

Search Terms: generic function currying ArgumentTypes infer

Code

type ArgumentTypes<F extends Function> = F extends (...args: infer A) => any
  ? A
  : never

const safeCurry = <
  F extends (...args: any[]) => any,
  G extends (fResult: ReturnType<F>) => any
>(
  gFn: G,
  fFn: F
) => (...args: ArgumentTypes<F>): ReturnType<G> => gFn(fFn(...args))


// (x:number) => number
const curried = safeCurry(
  (x: string) => parseInt(x),
  (x: number) => x.toString()
)

// (...args: any[]) => any
const curried2 = safeCurry(
  <T>(x: T) => x,
  (x: number) => x
)

Expected behavior:
the curried2 function should at least have the signature (x: number) => any, at best it would infer (x: number) => number

Actual behavior:
The args to the currried2 function are incorrect.

The reason why this is the case is confusing to me. Despite gFn being generic, and therefore harder for the compiler to understand what its ReturnType should be, the weird part is that this affects what fFn's argument type inference is. The type signature gFn should not affect how fFn is inferred.

Playground Link: playground link

Related Issues:

[andykais]

Found another way to express safeCurry using a third generic type. This correctly discovers the argument type of curried2 now, but the return type is still not inferable. Also, the type T is only inferred to be {} and therefore typechecking safeCurry's parameters no longer works (shown in the curried3 function).

type ArgumentTypes<F extends Function> = F extends (...args: infer A) => any
  ? A
  : never

const safeCurry = <
  T,
  F extends (...args: any[]) => T,
  G extends (fResult: T) => any
>(
  gFn: G,
  fFn: F
) => (...args: ArgumentTypes<F>): ReturnType<G> => gFn(fFn(...args))


// (x:number) => number
const curried = safeCurry(
  (x: string) => parseInt(x),
  (x: number) => x.toString()
)

// (x: number) => {}
const curried2 = safeCurry(
  <T>(x: T) => x,
  (x: number) => x
)

// this should throw an error, though it does not
// (x: number) => string
const curried3 = safeCurry(
  (x: string) => x,
  (x: number) => x
)

playground link

[RyanCavanaugh]

Tracking at #30134

[typescript-bot]

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