Conditional Type Produces undefined Unexpectedly

[m4dc4p]

TypeScript Version: 4.0.2

Search Terms:
conditional types never keyof required properties

Expected behavior:

In the code below, I would expect the type requiredKeysType1 to be "name" | "email".

Actual behavior:
The type of requiredKeysType1 is actually "name" | "email" | undefined. I had to define the Definable to filter out undefined and I don't understand why. The type requiredKeysType2 shows usage of that type.

My goal here is to define a const array of the required properties on a given interface for use in a type guard (isPerson). I would like the compiler to complain if I remove a key from the Person interface and forget to update the array.

Not implemented here but wished for - I would like the compiler to complain if I add a required property to Person and forget to update the array, but I can't figure out how to make that happen.

Related Issues:

Didn't find any related issues.

Code

interface Person {
  name: string
  age?: number
  email: string
}

type requiredKeysType1 = keyof RequiredProperties<Person>
// requiredKeys1 = "name" | "email" | undefined
const requiredKeys1: Array<requiredKeysType1> = ["name"]

type requiredKeysType2 = Definable<keyof RequiredProperties<Person>>
// requiredKeys2 = "name" | "email"
const requiredKeys2: ReadonlyArray<requiredKeysType2> = ["name"] as const

function isPerson2(s?: object): s is Person {
    return !! (s && requiredKeys2.every((k) => k in s))
}

/**
 * Remove undefined from the type given (most useful for unions)
 */
type Definable<T> = T extends undefined ? never : T

// Due to https://dev.to/busypeoples/notes-on-typescript-conditional-types-4bh
type RemoveUndefinable<Type> = {
  [Key in keyof Type]: undefined extends Type[Key] ? never : Key
}[keyof Type];

/**
 * Retrieves all the keys for the given object that are required (can't be
 * undefined.)
 */
type RequiredProperties<Type> = { // for some reason, keyof appends an `undefined` value. `Definable` filters that out.
  [Key in RemoveUndefinable<Type>]: Type[Key]
};

Output

"use strict";
// requiredKeys1 = "name" | "email" | undefined
const requiredKeys1 = ["name"];
// requiredKeys2 = "name" | "email"
const requiredKeys2 = ["name"];
function isPerson2(s) {
    return !!(s && requiredKeys2.every((k) => k in s));
}

Compiler Options

{
  "compilerOptions": {
    "noImplicitAny": true,
    "strictNullChecks": true,
    "strictFunctionTypes": true,
    "strictPropertyInitialization": true,
    "strictBindCallApply": true,
    "noImplicitThis": true,
    "noImplicitReturns": true,
    "alwaysStrict": true,
    "esModuleInterop": true,
    "declaration": true,
    "experimentalDecorators": true,
    "emitDecoratorMetadata": true,
    "moduleResolution": 2,
    "target": "ES2017",
    "jsx": "React",
    "module": "ESNext"
  }
}

Playground Link: Provided

[MartinJohns]

I'm confused. In the title you say it produces never, but then you don't mention it anywhere, neither in the actual behavior nor the expected behavior. The code doesn't produce a never either. Did you mean to write undefined in your title?

---

Duplicate of #34992.

You can fix your code by writing:

type RemoveUndefinable<Type> = {
  [Key in keyof Type]-?: undefined extends Type[Key] ? never : Key
}[keyof Type];

[m4dc4p]

Sorry about that, you are right. It produces "undefined" unexpectedly. Fixed. Thank for the reference to the fix & duplicate!

…

On Sat, Oct 17, 2020 at 12:45 AM Martin Johns ***@***.***> wrote: I'm confused. In the title you say it produces never, but then you don't mention it anywhere, neither in the actual behavior nor the expected behavior. The code doesn't produce a never either. Did you mean to write undefined in your title? — You are receiving this because you authored the thread. Reply to this email directly, view it on GitHub <#41145 (comment)>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AAAESAOPB2BLC4Y7B3QTSHTSLFDQLANCNFSM4ST6ZFSA> .