Retrieve the precise generic type from a function

Consider this scenario: There is a function called createObj that takes an object as a parameter of type ExampleType. It then creates a new object with a property def which contains the provided object. The goal is for it to return a type that matches the type of the provided object. Here's an example:

type ExampleType = { aaa: string };
declare type MyFunc = <T>(obj?: ExampleType) => { def: O };
const myFunc: MyFunc;

const a = myFunc({aaa: 'name'});

Currently, it returns { def: any }, but the desired output is { def: { aaa: 'name' } }. Is there a way to achieve this? Alternatively, consider this simpler example:

const aObj = { a: true };

type aType = { a: true };

Is there a way to make the type of aObj equal to aType?

EDIT: I have found a partial solution to my question (based on this Create TypeSafe Object literal based upon existing object in TypeScript).

function myFunc<O>(keysAndValuesObj: O) {
  return { def: keysAndValuesObj } as {
    def: { [K in keyof typeof keysAndValuesObj]: K };
  };
}
const a = myFunc({
  aaa: 'abc'
});
// now variable a has a type of { def: { aaa: 'abc' } }

Now, the next step is to ensure that the argument of myFunc is of type ExampleType. I attempted to extend O with the ExampleType type, but without success...

Answer №1

Based on my interpretation of your query, it seems like you are dealing with a function that accepts an object as an argument. Your objective is to produce an output object structured as { def: objectParemeter }, while retaining the original type of the input object.

If this is accurate, the following approach can be employed:

const createObject = <ObjectType extends object>(obj: ObjectType) => ({ def: obj })

If you wish to ensure that the return type exactly matches a specific type, you may consider:

const exampleResult = createObject<ExampleType>({ property: 'value' }) 

This will yield a type of { def: ExampleType }

You have the option to experiment further with this scenario here

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