为什么 Object.keys 在 TypeScript 中不返回 keyof 类型？

Question

Title says it all - why doesn't Object.keys(x) in TypeScript return the type Array<keyof typeof x> ? That's what Object.keys does, so it seems like an obvious oversight on the part of the TypeScript definition file authors to not make the return type simply be keyof T .

Should I log a bug on their GitHub repo, or just go ahead and send a PR to fix it for them?

Answer 1

The current return type ( string[] ) is intentional. Why?

Consider some type like this:

interface Point {
    x: number;
    y: number;
}

You write some code like this:

function fn(k: keyof Point) {
    if (k === "x") {
        console.log("X axis");
    } else if (k === "y") {
        console.log("Y axis");
    } else {
        throw new Error("This is impossible");
    }
}

Let's ask a question:

In a well-typed program, can a legal call to fn hit the error case?

The desired answer is, of course, "No". But what does this have to do with Object.keys ?

Now consider this other code:

interface NamedPoint extends Point {
    name: string;
}

const origin: NamedPoint = { name: "origin", x: 0, y: 0 };

Note that according to TypeScript's type system, all NamedPoint s are valid Point s.

Now let's write a little more code :

function doSomething(pt: Point) {
    for (const k of Object.keys(pt)) {
        // A valid call iff Object.keys(pt) returns (keyof Point)[]
        fn(k);
    }
}
// Throws an exception
doSomething(origin);

Our well-typed program just threw an exception!

Something went wrong here! By returning keyof T from Object.keys , we've violated the assumption that keyof T forms an exhaustive list, because having a reference to an object doesn't mean that the type of the reference isn't a supertype of the type of the value .

Basically, (at least) one of the following four things can't be true:

1. keyof T is an exhaustive list of the keys of T
2. A type with additional properties is always a subtype of its base type
3. It is legal to alias a subtype value by a supertype reference
4. Object.keys returns keyof T

Throwing away point 1 makes keyof nearly useless, because it implies that keyof Point might be some value that isn't "x" or "y" .

Throwing away point 2 completely destroys TypeScript's type system. Not an option.

Throwing away point 3 also completely destroys TypeScript's type system.

Throwing away point 4 is fine and makes you, the programmer, think about whether or not the object you're dealing with is possibly an alias for a subtype of the thing you think you have.

The "missing feature" to make this legal but not contradictory is Exact Types , which would allow you to declare a new kind of type that wasn't subject to point #2. If this feature existed, it would presumably be possible to make Object.keys return keyof T only for T s which were declared as exact .

---

Addendum: Surely generics, though?

Commentors have implied that Object.keys could safely return keyof T if the argument was a generic value. This is still wrong. Consider:

class Holder<T> {
    value: T;
    constructor(arg: T) {
        this.value = arg;
    }

    getKeys(): (keyof T)[] {
        // Proposed: This should be OK
        return Object.keys(this.value);
    }
}
const MyPoint = { name: "origin", x: 0, y: 0 };
const h = new Holder<{ x: number, y: number }>(MyPoint);
// Value 'name' inhabits variable of type 'x' | 'y'
const v: "x" | "y" = (h.getKeys())[0];

or this example, which doesn't even need any explicit type arguments:

function getKey<T>(x: T, y: T): keyof T {
    // Proposed: This should be OK
    return Object.keys(x)[0];
}
const obj1 = { name: "", x: 0, y: 0 };
const obj2 = { x: 0, y: 0 };
// Value "name" inhabits variable with type "x" | "y"
const s: "x" | "y" = getKey(obj1, obj2);

Answer 2

For a workaround in cases when you're confident that there aren't extra properties in the object you're working with, you can do this:

const obj = {a: 1, b: 2}
const objKeys = Object.keys(obj) as Array<keyof typeof obj>
// objKeys has type ("a" | "b")[]

You can extract this to a function if you like:

const getKeys = <T>(obj: T) => Object.keys(obj) as Array<keyof T>

const obj = {a: 1, b: 2}
const objKeys = getKeys(obj)
// objKeys has type ("a" | "b")[]

As a bonus, here's Object.entries , pulled from a GitHub issue with context on why this isn't the default :

type Entries<T> = {
  [K in keyof T]: [K, T[K]]
}[keyof T][]

function entries<T>(obj: T): Entries<T> {
  return Object.entries(obj) as any;
}

Answer 3

This is the top hit on google for this type of issue, so I wanted to share some help on moving forwards.

These methods were largely pulled from the long discussions on various issue pages which you can find links to in other answers/comment sections.

So, say you had some code like this :

const obj = {};
Object.keys(obj).forEach((key) => {
  obj[key]; // blatantly safe code that errors
});

Here are a few ways to move forwards:

1. If you don't need the keys and really just need the values, use .entries() or .values() instead of iterating over the keys.

    const obj = {}; Object.values(obj).forEach(value => value); Object.entries(obj).forEach([key, value] => value);

2. Create a helper function:

    function keysOf<T extends Object>(obj: T): Array<keyof T> { return Array.from(Object.keys(obj)) as any; } const obj = { a: 1; b: 2 }; keysOf(obj).forEach((key) => obj[key]); // type of key is "a" | "b"

3. Re-cast your type (this one helps a lot for not having to rewrite much code)

    const obj = {}; Object.keys(obj).forEach((_key) => { const key = _key as keyof typeof obj; obj[key]; });

Which one of these is the most painless is largely up to your own project.

Answer 4

I'll try to explain why object keys cannot return keyof T while being safe with some simple example

// we declare base interface
interface Point {
  x: number;
  y: number;
}

// we declare some util function that expects point and iterates over keys
function getPointVelocity(point: Point): number {
  let velocity = 0;
  Object.keys(point).every(key => {
    // it seems Object.keys(point) will be ['x', 'y'], but it's not guaranteed to be true! (see below)
    // let's assume key is keyof Point
    velocity+= point[key];
  });

  return velocity;
}

// we create supertype of point
interface NamedPoint extends Point {
  name: string;
}

function someProcessing() {
  const namedPoint: NamedPoint = {
    x: 5,
    y: 3,
    name: 'mypoint'
  }

  // ts is not complaining as namedpoint is supertype of point
  // this util function is using object.keys which will return (['x', 'y', 'name']) under the hood
  const velocity = getPointVelocity(namedPoint);
  // !!! velocity will be string '8mypoint' (5+3+'mypoint') while TS thinks it's a number
}

Answer 5

Possible solution

const isName = <W extends string, T extends Record<W, any>>(obj: T) =>
  (name: string): name is keyof T & W =>
    obj.hasOwnProperty(name);

const keys = Object.keys(x).filter(isName(x));

Answer 6

In loops to get around this:

  let key2: YourType
  for (key2 in obj) {
    obj[key2]
  }