Optional Generic Type Inference

#13487 added default generic types, but it's still not possible to infer a generic type:

type Return<T extends () => S, S = any> = S

Here S takes its default type any, but T could permit inference of a subset type of any for S:

const Hello = () => 'World'

type HelloReturn = Return<typeof Hello> // any

Here HelloReturn still has any type, and TypeScript could infer S as the literal type 'World', which is a subset of any.

Use Case

Here's an example of a fully declarative workaround for #6606, using generic type inference:

type Return<T extends () => S, S = any> = S

const Hello = () => 'World'

type HelloReturn = Return<typeof Hello>  // 'World'

Default Type

If return of T is not a subset of S, it should throw an error:

type ReturnString<T extends () => S, S = string> = S

const Hello = () => 'World'
type HelloReturn = ReturnString<typeof Hello>  // 'World'

const calculateMeaningOfLife = () => 42
type MeaningOfLife = ReturnString<typeof calculateMeaningOfLife> // ERROR: T should return a subset of String

Multiple Possible Type Inferences

The current implementation always returns the superset (which is just the default generic type), solving this issue would require to return the subset (the most precise type of all the inferred possibilities).

If a type has multiple possible type inferences, TypeScript should check that all these types are not disjoint, and that they all are subsets of the Default Generic Type.

The inferred type is the most precise subset.

The issue names no files, tests, or entry points. Start by reproducing the generic-inference examples in a TypeScript type-checking environment and tracing the existing behavior for default generic types; done means the inferred type is the most precise valid subset and incompatible returns are rejected.