Narrow type after assertion?

[aaronbeall]

TypeScript Version: 2.7.0-dev.201xxxxx

Would it be possible to treat an assertion like a type-guard and narrow the type within blocks that are "guarded" by an assertion?

For example:

Code

type Square = { size: number; }
type Circle = { radius: number; }
type Shape = Square | Circle;

declare const shape: Shape;
if ((shape as Circle).radius) {
  shape.radius // Error: Property 'radius' does not exist on type 'Square'
  (shape as Circle).radius // Ok, but seems redundant
}

But you could consider the (shape as Circle).radius like the user saying:

isCircle(shape) && shape.radius
// or at least
hasRadius(shape) && shape.radius

I realize that checking for a property is not considered good enough to narrow the type, but if the user has explicitly asserted the type it would be convenient if control flow could keep track of that in downstream code.

[RyanCavanaugh]

Same as #8655?

[aaronbeall]

@RyanCavanaugh Looks like the same desired behavior for a different use case (using an assert() function)... in this case I'm saying can we treat the existing assertion syntax as a type-guard?

[aaronbeall]

I realize that checking for a property is not considered good enough to narrow the type

Oh, actually, I just saw that #10485 "Treat in operator as type guard" made the roadmap for 2.7! I think that gives me everything I wanted here?

if ("radius" in shape && shape.radius) {
  shape.radius // Ok?
}

[typescript-bot]

Automatically closing this issue for housekeeping purposes. The issue labels indicate that it is unactionable at the moment or has already been addressed.