What steps do I need to take for the function to accurately determine the return type?

class Foo {
  name: string;

  constructor({name}: {name: string}) {
    this.name = name;
  }
}

class Bar<T extends Foo> {
  foo: T;

  constructor({foo}: {foo: T}) {
    this.foo = foo;
  }
}

class CustomFoo extends Foo {
  xxx: string;

  constructor({
    name,
    xxx,
  }: {
    name: string,
    xxx: string,
  }) {
    super({name});
    this.xxx = xxx;
  }
}

class CustomBar<F extends Foo> extends Bar<F> {
  customField: string;

  constructor({
    foo,
    customField,
  }: {
    foo: F,
    customField: string,
  }) {
    super({foo});
    this.customField = customField;
  }
}

const doSomething = <
  B extends Bar<F>,
  F extends Foo,
>(
  FooConstructor: { new(...args : any[]): F; },
  BarConstructor: { new(...args : any[]): B; },
): B => {
  return new BarConstructor({});
}

const mything = doSomething(CustomFoo, CustomBar);

Unfortunately the type of mything is Bar<CustomFoo> instead of CustomBar<CustomFoo>.

It seems TypeScript fails to infer the correct return type.

How can I modify doSomething to return the expected inferred type?

I am unsure about potential solutions to this issue.

Answer №1

Unfortunately, TypeScript's inference does not give the desired outcome, and there is no straightforward way in TypeScript to express what you're trying to achieve seamlessly or broadly. TypeScript lacks support for "call types" as described in microsoft/TypeScript#40179. Ideally, one would want to deduce the type returned when calling a constructor based on its generic input or invocation method, but without directly invoking the constructor, this information remains inaccessible. While conditional types like InstanceType can be employed, they do not preserve any generic specifics. The core issue arises from higher-kinded types, which are absent in TypeScript, as requested in microsoft/TypeScript#1213. Managing generics within other generics poses a considerable challenge.

For example, when

doSomething(CustomFoo, CustomBar)
is invoked, F is inferred as CustomFoo, and then B must be inferred as a subtype of Bar<CustomFoo>. However, TypeScript struggles at this point to incorporate CustomFoo due to the lack of higher-kinded types. The best outcome attainable is CustomBar<Foo>, yet TypeScript defaults to the constraint of
Bar<CustomFoo></code. Consequently, in this context, TypeScript only infers either <code>CustomBar<Foo>
or Bar<CustomFoo>.


To tackle this limitation, one workaround involves utilizing an instantiation expression with CustomBar to inform the compiler that you specifically intend to insert CustomFoo into the type argument for that constructor:

const mything = doSomething(CustomFoo, CustomBar<CustomFoo>);
// const mything: CustomBar<CustomFoo>

By doing this, when the compiler deduces B, the only viable choice becomes CustomBar<CustomFoo>,

Playground link to code

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