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

There is an existing feature request at microsoft/TypeScript#12184 to allow such type guard results to be "saved" into a named value to be used later. The request is open but listed as "revisit" because there's no obvious way to implement it in a performant way. The word from the lead architect of the language is:

This would require us to track what effects a particular value for one variable implies for other variables, which would add a good deal of complexity (and associated performance penalty) to the control flow analyzer. Still, we'll keep it as a suggestion.

So I wouldn't expect to see such a feature in the language anytime soon, unfortunately.

---

My suggestion is just to continue using your inline type check. If you have a more complex type guard situation it may be worthwhile to make a user-defined type guard function but I don't see that as an improvement for the case in your example code.

---

Okay, hope that helps; good luck!

Answer 2

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
  }
};