Determination of function return type based on the presence of an optional argument

I am attempting to design a function that will have a return type based on whether an optional parameter (factory) is provided or not

  • If the factory parameter is passed, then the return type should match the return type of the factory
  • Otherwise, it should default to 'any'

This is how I tried to implement it:

type Factory<T> = (data: any) => T;

function getObject<T>(id: number, factory?: Factory<T>): Factory<T> extends undefined ? any : ReturnType<Factory<T>> {
  let obj; // some entity from db in real app
  return factory ? factory(obj) : obj;
}

getObject(1, () => new Date()).getDate();             // Works fine, treated as a date
getObject(1, () => new String()).toLocaleLowerCase(); // Correctly recognized as a string
getObject(1).anything; // This results in an error, TypeScript interprets it as {} instead of any 

However, when I do not provide the factory parameter, it does not behave as expected. What changes can I make to fix this? https://stackblitz.com/edit/typescript-xci2gs

Answer №1

I successfully resolved the issue by updating the signature to

function getObject<T extends Factory<any>>(id: number, factory?: T): T extends undefined ? any : ReturnType<T> {

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