Can TypeScript infer type of a discriminated union via “extracted” boolean logic?

Question

I have been using discriminated unions (DU) more frequently and have come to love them. I do however have one issue I can\'t seem to get past. If you inline a boolean check fo

Answer 1

Currently nope as mentioned by @jcalz, however there's a minimal workaround that you can always apply to overcome this problem.

The trick is to make extractCheck a function that returns type predicate.
(Refer https://www.typescriptlang.org/docs/handbook/advanced-types.html#user-defined-type-guards)

The following is a working example:

type A = { type: "A"; foo: number };
type B = { type: "B"; bar: string };
type DU = A | B;

const checkIt = (it: DU) => {
  // Make `extractCheck` a function 
  const extractedCheck = (it: DU): it is A => it.type === "A";

  if (extractedCheck(it)) {
    console.log(it.foo);  // No error
  }
};