`void | T` allows narrowing of `undefined | T`

[robbiespeed]

`void | T` allows narrowing of `undefined | T`

[MartinJohns]

Related: #42709

In general a union with void makes absolutely no sense. Unfortunately a lot of people think void means "no value", but that's not the case.

[RyanCavanaugh]

There's #46837 then

[fatcerberus]

I still get a kick out of this comment

We've spent several dozen person-hours in the past few months beating this particular horse to death without much forward progress, but I remain optimistic that we'll eventually make some progress on it.

Given that that was posted over 3 years ago, I think it's safe to say the horse is well and truly dead by now and any further progress on that front will almost assuredly require enlisting the services of a necromancer

[robbiespeed]

@MartinJohns Thanks for pointing me to that issue I was tempted to make a feature request for at the very least inferring function returns as undefined by default, if not outright removing void in strict mode.

I do agree that in general void shouldn't be used with a union, but it illustrates the inconsistent behaviour of void well. An example without a union is:

declare const v: void;
if (v !== undefined) {
  v; // type is never (void narrowed as if it was undefined)
} else {
  v; // void?
}

[RyanCavanaugh]

I think #42709 is the controlling issue here - removing all conflations of void and undefined is really the knot to untangle

[robbiespeed]

@RyanCavanaugh I don't agree, that issue seems to be centered around replacing void with undefined as the default inferred value for non returning functions, but still allowing explicit use of void.

This specific bug will still be present regardless of the decision there.

[robbiespeed]

Just saw that as well, will close this.