Return different type by string in Typescript

Question

Given we have two different types as below, how can we change the return of function based on a string parameter and not providing the generic type?

interface Type1 { typeName: string; };
interface Type2 { typeVersion: string; };
type AllTypes = Type1 | Type2;

function returnTypes(t: string): AllTypes {
  ...
}

const typeResult = returnTypes('type1');
console.log(typeResult.typeName);

Here the return is not defined!

Answer 1

Create an overloaded declaration like so

interface Type1 { typeName: string; };
interface Type2 { typeVersion: string; };
type AllTypes = Type1 | Type2;

function returnTypes(t: 'type1'): Type1;
function returnTypes(t: 'type2'): Type2;
function returnTypes(t : 'type1' | 'type2'): AllTypes {
    switch (t) {
        case "type1": return { typeName: "test" };
        case "type2": return { typeVersion: "test" };
    }
}

console.log(returnTypes('type1').typeName);
console.log(returnTypes('type2').typeVersion);

Note that when you overload a function in this manner, the implementation signature is not available to callers. Only the declarations specified beforehand comprise an overloaded function's interface.

UPDATE: Fixed and completed example to show that TypeScript knows which type it is returning.