Can typescript check every Record value of Enum?

Question

Typescript is beautiful, and I love to write on typescript cause I can just write

enum Test {
    some = 'some',
    another = 'another',
}

const test: Record<Test, string> = {
    some: 'string'
}

then typescript asks me to type every key value from Test as the key of my test object

Property 'another' is missing in type '{ some: string; }' but required in type 'Record<Test, string>'

But if I want to write an opposite operation (like parse/stringify), I want to be sure that I don't miss some Test as values of Record

enum Test {
    some = 'some',
    another = 'another',
}
const test: Record<string, Test> = {
    some: Test.some
}

But typescript assume that it is correct.

I would like to throw some error by typescript.

Is there any way in typescript to check every Record value of Enum?

Answer 1

TypeScript doesn't really have the idea of "exhaustive property types", but you could write a generic type along with a helper function to validate that a given value behaves as you like. For example:

const exhaustive = <V,>() => <T extends Record<keyof T, V>>(
    t: { [K in keyof T]: [V] extends [T[keyof T]] ? T[K] : Exclude<V, T[keyof T]> }
) => t;

This is a curried function where you manually specify the value type V you'd like to "exhaust", and it outputs another generic function that does the check:

enum Test {
    some = 'some',
    another = 'another',
}   
const exhaustiveTest = exhaustive<Test>()

(The reason you need to use two functions like that is because TypeScript doesn't let you manually specify a generic type argument while inferring another one; there is an open feature request for this at microsoft/TypeScript#26242 and the various workarounds including currying are discussed in Typescript: infer type of generic after optional first generic )

---

When you call exhaustiveTest() it will return its input, but a compiler error will occur if you are missing any of the values in Test :

const test = exhaustiveTest({
    abc: Test.some,
    def: Test.another
}); // okay

const badTest = exhaustiveTest({
    abc: Test.some // error! Type 'Test.some' is not assignable to type 'Test.another'
})

Hooray, that's what you wanted!

---

The way it works is that the call to exhaustiveTest() infers the type argument T to be the type passed in as the function argument t , but it checks it against the mapped type { [K in keyof T]: [Test] extends [T[keyof T]]? T[K]: Exclude<Test, T[keyof T]> } { [K in keyof T]: [Test] extends [T[keyof T]]? T[K]: Exclude<Test, T[keyof T]> } . Each property is the conditional type [Test] extends [T[keyof T]]? T[K]: Exclude<Test, T[keyof T]> [Test] extends [T[keyof T]]? T[K]: Exclude<Test, T[keyof T]> . That means: if every value of type Test can be assigned to one of the property types of T , then each property of T is just fine as itself (the indexed access type T[K] ); otherwise, it should be Exclude<Test, T[keyof T]> using the Exclude<T, U> utility type to filter out all the properties of T from Test to get just the missed values, so that the error message will mention these missing values.

As a concrete case, if T is {abc: Test.some; def: Test.another} {abc: Test.some; def: Test.another} , then T[keyof T] is Test.some | Test.another Test.some | Test.another , which is the same as Test , and so the mapped type is just {abc: Test.some; def: Test.another} {abc: Test.some; def: Test.another} , and it type checks. But if T is just {abc: Test.some} , then T[keyof T] is just Test.some , which is missing thing from Test . And so the mapped type is {abc: Test.another} , and it fails to type check, complaining about the missing thing.

Playground link to code