"in" operator not narrowing the type if used with a const

[kdmadej]

Bug Report

🔎 Search Terms

• narrow in operator
• in operator
• narrow operator
• operator type guard

🕗 Version & Regression Information

This is the behavior in every version I tried, and I reviewed the FAQ for entries about the "in" operator.

Tested on all TS versions available in the playground (at the time of writing: 3.3 - Nightly of 4.2.0)

⏯ Playground Link

Playground link with relevant code

💻 Code

const keywordA = 'a'
const keywordB = 'b'

type A = { [keywordA]: number }
type B = { [keywordB]: string }

declare const c: A | B

if ('a' in c) {
    c // narrowed to `A`
}

if (keywordA in c) {
    c // not narrowed, still `A | B`
}

🙁 Actual behavior

When using a const string as the "prop" argument for the in operator the type of the object is not narrowed.

🙂 Expected behavior

Since I'm using a const, thus dealing with a string literal type (same as when using a literal value as an argument), I'd expect the type to be narrowed inside the if block.

[RyanCavanaugh]

I don't see any reason this couldn't be made to work