Type narrowing not working correctly with Uppercase<string> & Lowercase<string>

[wartab]

🔎 Search Terms

"narrowing lowercase" "narrowing uppercase"

🕗 Version & Regression Information

• This is the behavior in every version I tried, and I reviewed the FAQ for entries about Common Bugs and Type System Behavior

⏯ Playground Link

https://www.typescriptlang.org/play/?#code/DYUwLgBAxg9grgOzAJwJ4GEYBMQC4IBEAQgKIEQA+hAMgKrlUEBy1BA3AFAewIDOkAB2QgAZgEsAHhAC8hAEYgCAOjAxaAgSGToAhrxAAKAJQQ9EdZuRQ9IADz9kYhAHMAfJw5iREA0NGSZaVliMkoqP3EpIOC6BnDhSMDglgITAG8OCCzoeCQ0TBwZCAjJTgB6MqyAPQ4KrIAVVE0IAHILLWt9exQnNxaIMV4IBBhIPV4xZwQdOVAIVXmmkFaQhhp6SkIUlqVayohG5paHXv7B4dHTXgmpmbmFsCWV0jWCWM3mVh2OAF8gA

💻 Code

let countryCode: "BE" | "LU" | "NL";

const prefix = "be".toUpperCase() as Uppercase<string>;

if (prefix === "BE" || prefix === "LU" || prefix === "NL") {
    countryCode = prefix;
//  ^
//  Type 'Uppercase<string>' is not assignable to type '"BE" | "LU" | "NL"'.
//  Type 'string' is not assignable to type '"BE" | "LU" | "NL"'.
}

🙁 Actual behavior

TypeScript cannot narrow down an Uppercase<string> or Lowercase<string> type after explicitly comparing against specific values.

🙂 Expected behavior

TypeScript should accept the code above as "BE" | "LU" | "NL" is strictly a sub-type of Uppercase<string>.

Additional information about the issue

No response

[alex-vukov]

It doesn't look like a bug to me. You are trying to fit a wider type inside a narrower type. The type of prefix is Uppercase<string> which means any uppercase string. The countryCode is "BE" | "LU" | "NL". You can't fit any uppercase string into something which accepts only the three possible strings hence the error. If you remove the as Uppercase<string> cast it works correctly because now it can be narrowed automatically.

[wartab]

Considering "lu" satisfies Uppercase<string> accurately doesn't compile and "LU" satisfies Uppercase<string> does compile, I believe that TypeScript should have sufficient info to do the same type narrowing with Uppercase<string> as with string.

If you remove as Uppercase<string>, prefix just becomes just string, which is a wider type than Uppercase<string>, since you can pass it to anything that is just a string.

So this does look like a bug to me.

[alex-vukov]

I see what you mean but when you declare let countryCode: "BE" | "LU" | "NL"; the TS parser knows that "BE", "LU, and "NL" are strings because they are delimited by quotes. However, for the TS parser to know if they are uppercase, lowercase, starting with a capital letter, etc... it needs to parse each of them in several different ways which can be very ineffective especially if they are many. When you do "LU" satisfies Uppercase<string> the parser will just check the LU string to see if it's uppercase so it's a single check for a single string so it's simple.

[MartinJohns]

The actual issue is that the compiler can't narrow down Uppercase<string> to Uppercase<"BE"> when comparing to such a string. I don't know if there's an open issue for this specifically.

[jcalz]

This is at the very least a missing feature (if not a bug). One might expect that Uppercase<string> could be narrowed to an uppercase string literal via equality check. If you write a user-defined type guard then it works as expected:

const isEqual = <T, U extends T>(t: T, u: U): t is U => t === u

if (isEqual(prefix, "BE") || isEqual(prefix, "LU") || isEqual(prefix, "NL")) {
    countryCode = prefix; // okay
}

Playground link

[RyanCavanaugh]

It's also interesting that the comparability relation doesn't properly account for this

function fn(foo: Uppercase<string>) {
    // No error, ?
    if (foo === "ba") {}
}

fn("boo"); // error
fn("BOO"); // not error