how to return generic object that extends class in typescript method?

Question

I am trying to understand how to use generic in typescript. I have the following code

  foo<T extends X>(a:A,b:B): T{
    // some logic 
    const c: C = doSomthing(a,b)
    if (c!= null){
      return {
        a: A,
        b: B,
        c: C
      } as Y; // Y extends X
    } else {
      return {
        a: A,
        b: B,
      } as X;
    }
  }

However, it won't work: Y is assignable to the constraint of type T, but T could be instantiated with a different subtype of constraint How can I implement it right?

Answer 1

T extends X means all types which have at least members a: A and b: B. Your function can't create objects satisfying every one of those types.

A correct typing for your function is X. It isn't a generic function.

function foo(a:A,b:B): X {
  // some logic 
  const c: C | null = doSomthing(a,b)
  return c != null ? { a, b, c } as Y : { a, b };  
}

You could also use the type X | Y, which removes the need for the as Y. X | Y has the same values as X, so is in some sense equivalent.

N.b. it is incorrect to specify the type of elements in an object literal, as { a: A } is the syntax for associating the key a with the value A.