CODEWITHSUNDEEP
typescripttypesreduce
marvinhagemeister
marvinhagemeister

Reputation: 3144

TypeScript intersection types with array reduce

I'm trying to get typings to work with a generic array reduce function which basically merges two objects. The following snippet is a dumped down version of the real code. Why is fl's type {} and not IFoo & IBar?

(I'm aware that this particular example can easily be replaced with a single Object.assign() call.)

const flatten = <K, T>(prev: K, x: T): K & T => {
  return Object.assign(prev, x);
};

interface IFoo {
  foo: true;
}

interface IBar {
  bar: true;
}

const fooRes: IFoo = { foo: true };
const barRes: IBar = { bar: true };

const fl = [fooRes, barRes].reduce(flatten, {});

console.log(fl); // here, fl : {}

Upvotes: 3

Views: 5676

Answers (1)

Benjamin Hodgson
Benjamin Hodgson

Reputation: 44634

The signature of reduce is

reduce<U>(callbackfn: (previousValue: U, currentValue: T, currentIndex: number, array: T[]) => U, initialValue: U): U

(T is the type parameter of the array itself.) So, faced with the code

[fooRes, barRes].reduce(flatten, {})

the type checker's job is to figure out what U is. Let's walk through its reasoning:

  1. fooRes : IFoo and barRes : IBar, so [fooRes, barRes] : (IFoo | IBar)[]
  2. Thus, the array's T ~ IFoo | IBar
  3. So flatten is being called with its T parameter set to IFoo | IBar
  4. flatten's return type (K & T) is therefore K & (IFoo | IBar)
  5. Since flatten's return type must be assignable to U, that gives us the constraint U >= (U & (IFoo | IBar)), which simplifies to U >= (IFoo | IBar)
  6. The other bit of evidence is the initialValue parameter which has a type of {}
  7. So U >= {}
  8. The least upper bound of these two constraints is {}. So the type checker infers U ~ {}.

Why doesn't it realise that the return type is IFoo & IBar? The type checker doesn't reason about the runtime behaviour of your code - that flatten's parameter takes on a variety of different types throughout the reduction. An array of type (IFoo | IBar)[] is not guaranteed to have both IFoos and IBars in it - it could just be an array of IFoos. Deducing that flattening an heterogeneous list squashes its constitutent types down would require quite a sophisticated proof, and it doesn't seem reasonable to expect a machine to be able to write such a proof for you.

Upvotes: 11

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