Refining one value in a union does not refine all values

[ghost]

TypeScript Version: 3.3

Search Terms:

refine union

Code

declare function foo(): [number, string] | [string, number];
const [a, b] = foo();

if (typeof a === 'number') {
  console.log(a, b);
}

Expected behavior:

Would expect b to be of type string here

Actual behavior:

a is refined to number, while b remains string | number

Playground Link:

link

[jack-williams]

There are two issues at play here:

1. Narrowings are not tracked across variables like this. The issue for this is: Indirect type narrowing via const #12184

2. A typeof check will not narrow a union like a discriminant property check, it will only narrow the type of the checked property. This has been the design since CFA was introduced I think, but I don't know if there are any plans to revisit this.

[ghost]

Is there a different way to either define the return type or do the refinement so that this will work without needing to refine each? Also tried leaving the entire object in tact but that doesn't work either for some reason:

declare function foo(): [number, string] | [string, number];
const res = foo();

if (typeof res[0] === 'number') {
  console.log(res); // [number, string] | [string, number]
}

[typescript-bot]

This issue has been marked as a 'Duplicate' and has seen no recent activity. It has been automatically closed for house-keeping purposes.