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The evaluation of the Keyof Type Operator produces different results to the Pick built-in tool. #48173
Description
Bug Report
🔎 Search Terms
Pick
🕗 Version & Regression Information
v4.5.4
⏯ Playground Link
💻 Code
interface User {
name?: string
age?: number
address?: string
}
type Foo<T, U = keyof T> = {
[P in Extract<U, keyof T>]: T[P]
}
type Bar<T,U=keyof T> = Required<Pick<T, Extract<U, keyof T>>>
type Res = Foo<User, 'name'> //name: string | undefined;
type Res2 = Bar<User, 'name'> //name: string
// We can quickly address your report if:
// - The code sample is short. Nearly all TypeScript bugs can be demonstrated in 20-30 lines of code!
// - It doesn't use external libraries. These are often issues with the type definitions rather than TypeScript bugs.
// - The incorrectness of the behavior is readily apparent from reading the sample.
// Reports are slower to investigate if:
// - We have to pare too much extraneous code.
// - We have to clone a large repo and validate that the problem isn't elsewhere.
// - The sample is confusing or doesn't clearly demonstrate what's wrong.🙁 Actual behavior
type Res = Foo<User, 'name'> //name: string | undefined;
The value evaluated for the name contains undefined
🙂 Expected behavior
type Foo<T, U = keyof T> = {
[P in Extract<U, keyof T>]: T[P]
}
[P in Extract<U, keyof T>]: T[P] should not return undefined because Extract only narrows down the range of the keys
Activity
jcalz commentedon Mar 8, 2022
I don’t see a bug here. The type
User["name"]isstring | undefined, so that’s the type of thenameproperty ofFoo<User, "name">. Homomorphic mapped types and mapped type modifiers can complicate this, but neither of those are happening inFoo.RyanCavanaugh commentedon Mar 8, 2022
+1 to jcalz; this behaves exactly the same as a step-by-step manual evaluation of the type
someBrown commentedon Mar 8, 2022
new playground link
Sorry, I forgot to add the '-?' operator in type Foo. I changed the name attribute to required, why does it still return undefined?
fatcerberus commentedon Mar 8, 2022
User["name"]evaluates tostring | undefinedbecause the property is optional.-?removes optionality in the new type, but this is separate from the evaluation ofT[P], so you end up withname: string | undefined. You might considerNonNullable<T[P]>.ben-lau commentedon Mar 9, 2022
But there is another question: I think the
?symbol means this property is optional. So in this case, thenamewhich is the keyofUsersshould be optional, but if you got this property , it has to bestringnotundefinedas well.jcalz commentedon Mar 9, 2022
If you have a value
uof typeUser, then inconst n = u.name, the type ofnisstring | undefined. That's just how it works. Whether or not the key is present vs absent and whether or not you can actually writeundefinedtou.namehas to do with--exactOptionalPropertyTypes(#43947) but that doesn't change things here.ben-lau commentedon Mar 9, 2022
All right, I missed this, thanks.