Checking an optional generic parameter for undefined results in T & ({} | null)

caTS 0
Viewed 46

So I was answering this question just now, and in it I found something strange that I could not find explanations of.

If we have a generic function, and the parameter is optional and the parameter's type is using the generic, something like this:

function ok<T>(thing?: T) {

}

Then when we try to narrow thing inside the function to just T explicitly checking if it is undefined:

thing !== undefined ? thing : 0;
//                    ^? T & ({} | null)

Then thing is narrowed to T & ({} | null). Moreover, when we use typeof, this is narrowed to the expected T. This also behaves the same with if statements.

typeof thing !== "undefined" ? thing : 0;
//                             ^? T

Here's a minimal reproducible example with all the things I mentioned:

function ok<T>(thing?: T) {
    thing !== undefined ? thing : 0;
    //                    ^?

    if (thing !== undefined) {
        thing
    //  ^?
    }

    typeof thing !== "undefined" ? thing : 0;
    //                             ^?

    if (typeof thing !== "undefined") {
        thing
    //  ^?
    }
}

Why is this the case? Do the two conditions actually have different behavior that I didn't know about? Is this a bug? I'm looking for an explanation for this behavior.

typescript
2 Answers

Very interesting questions.

I think the reason is something like this:

  • according to this when using an unconstrained generic T could match both
    • null and
    • {}
  • this will lead you do have a type similar to T | null | {}.
  • from this it makes sense that when doing T !== undefined to end up with something similar to T & ({} | null).
  • but when doing a typeof to get different results

Also. You can check 2 different things

  • Add a constraint to the generic. Something like this function ok<T extends string>. This will lead to consistent results.
  • Or use != instead of !== which will turn thing in NonNullable<T>. As it can be seen here

It seems like a long, long time ago {} was the top type for generics and changing it now would break a lot of existing code.

Radu Diță0

This is the result of improved intersection reduction, union compatibility, and narrowing, as implemented in microsoft/TypeScript#49119 and released in TypeScript 4.8.

Certain type guards will now narrow formerly un-narrowable (is that a word?) types by intersecting them with a filtered version of {} | null | undefined, a union type equivalent to the unknown type because the empty object type {} accepts all values except null and undefined (see How to undestand relations between types any, unknown, {} and between them and other types? for more info).

Specifically, the PR mentions:

  • In control flow analysis of equality comparisons with null or undefined, generic types are intersected with {}, {} | null, or {} | undefined in the false branch.

In your case, the value thing started off as type T | undefined, and once you eliminate undefined from that with an equality check, you get T & ({} | null) which more accurately represents the fact that thing cannot be undefined than T alone would. While inference would usually prevent this for your exact example, nothing stops someone from writing ok<string | undefined>(Math.random()<0.5 ? "abc" : undefined), and so the type T would be string | undefined whereas T & ({} | null) is just string (not string & {}; see the PR for details about why and how this is different).

On the other hand, this logic was not implemented for the conceptually equivalent typeof thing !== "undefined" check. The only typeof check I see being affected here is for typeof xxx === "object". Whether or not they would consider the difference in narrowing between x === undefined and typeof x === "undefined" to be a bug, design limitation, intentional, or a missing feature isn't documented there, but TypeScript is definitely behaving as designed. Maybe it would be worthwhile to open a new issue or at least comment about it in GitHub? Not sure.

jcalz0
Related