Enable type parameter lower-bound syntax

[jyuhuan]

TypeScript Version: 2.1.1

Code

class Animal {}
class Cat extends Animal {}
class Kitten extends Cat{}

function foo<A super Kitten>(a: A) { /* */ }

Expected behavior:
The type parameter A has the type Kitten as lower-bound.

Actual behavior:
Compilation failure. The syntax is unsupported.

Discussion:
The upper-bound counterpart of the failed code works fine:

class Animal {}
class Cat extends Animal {}
class Kitten extends Cat{}

function foo<A extends Animal>(a: A) { /* */ }

People in issue #13337 have suggested to use

function foo <X extends Y, Y>(y: Y) { /* */ }

to lower-bound Y with X. But this does not cover the case where X is an actual type (instead of a type parameter).

[RyanCavanaugh]

What is this useful for?

[jyuhuan]

@RyanCavanaugh : In short, it mimics contravariance, just as extends mimics covariance.

We will try to sort an array of cats to see the necessity of this feature.

To do comparison-based sorting, we need a Comparator interface. For this example, we define it as follows:

interface Comparator<T> {
  compare(x: T, y: T): number
}

The following code shows that the class Cat has Animal as its super-class:

class Animal {}
class Cat extends Animal {}

Now we can write a sorting function that supports arbitrary Cat comparators as follows:

function sort(cats: Cat[], comparator: Comparator<Cat>): void {
  // Some comparison-based sorting algorithm.
  // The following line uses the comparator to compare two cats.
  comparator.compare(cats[0], cats[1]);
  // ...
}

Now, we will try to use the sort function. The first thing is to implement a CatComparator:

class CatComparator implements Comparator<Cat> {
  compare(x: Cat, y: Cat): number {
    throw new Error('Method not implemented.');
  }
}

Then we create a list of Cats,

const cats = [ new Cat(), new Cat(), new Cat() ]

Now we can call sort as follows without any problem:

sort(cats, new CatComparator());

We have not seen the need for contravariance so far.

Now, suppose we are told that someone has already implemented a comparator for Animals as follows:

class AnimalComparator implements Comparator<Animal> {
  compare(x: Animal, y: Animal): number {
    throw new Error('Method not implemented.');
  }
}

Since a Cat is also an Animal, this AnimalComparator is also able to handle Cats, because the compare function in AnimalComparator takes two Animals as input. I can just pass two Cats to it and there will be no problem.

Naturally, we would want to use AnimalComparator for sort too, i.e., call the sort function as:

sort(cats, new AnimalComparator());

However, since the following two types:

• Comparator<Animal>
• Comparator<Cat>

are not related from the point of view of TypeScript's type system, we cannot do that.

Therefore, I would like the sort function to look like the following

function sort<T super Cat>(cats: Cat[], comparator: Comparator<T>): void {
  // Some comparison-based sorting algorithm.
  // The following line uses the comparator to compare two cats.
  comparator.compare(cats[0], cats[1]);
  // ...
}

or as in Java,

function sort(cats: Cat[], comparator: Comparator<? super Cat>): void {
  // Some comparison-based sorting algorithm.
  // The following line uses the comparator to compare two cats.
  comparator.compare(cats[0], cats[1]);
  // ...
}

I am aware of the fact that TypeScript does not complain if I pass AnimalComparator to sort. But I would like TypeScript's type system to explicitly handle type lower-bounds. In fact, the current type system of TypeScript will let some type error closely related to this issue silently pass the compiler's check (see issue #14524).

[jklmli]

This is the best example of contravariance I've read in a long time. Props! 🙌

[btipling]

As a reference point, flowtype uses - and + to indicate covariance and contravariance

class ReadOnlyMap<K, +V> {
  store: { +[k:K]: V };
  constructor(store) { this.store = store; }
  get(k: K): ?V { return this.store[k]; }
}

[HerringtonDarkholme]

Hi @jyuhuan , since you probably already know this https://github.com/Microsoft/TypeScript/wiki/FAQ#why-are-function-parameters-bivariant

I'm afraid lower bound isn't that useful in current TypeScript where variance is unsound.

Indeed, there are also cases lower bound can be useful without variance. Like

interface Array<T> {
  concat<U super T>(arg: ): Array<U>
}

var a = [new Cat]
a.concat(new Animal) // inferred as Animal[]

In such case like immutable sequence container, lower bound helps TypeScript to enable pattern where new generic type is wider than original type.

Yet such usage still needs more proposal since TS also has union type. For example should strArr.concat(num) yield a Array<string | number>?

[zpdDG4gta8XKpMCd]

migrated from #14728:

currently when we see a constraint <A extends B> it means A is a subtype of B, so that

declare var a: A;
declare var b: B;
b = a; // allowed
a = b; // not allowed

consider adding a new constraint of the reversed relation: <A within B> that would mean A is a supertype of B (or in other words B is subtype of A), so that:

declare var a: A;
declare var b: B;
b = a; // not allowed
a = b; // allowed

use case

i have a BigFatClass with 50 methods and a 1 little property, now i want to run some assertions over it, if i declare these assertions like expect<T>() and toEqual<T> of the same T then toEqual asks me for 50 methods that don't matter for the test, and that's the problem

what i need it to declare expect<T>() and toEqual<U within T>() so that i could simply write:

expect(new BigFatClass()).toEqual({ value: true });

[aaronbeall]

<U within keyof T> could be confusing, since U in this case should be a super-set of T keys, not "within" the keys of T.

[zpdDG4gta8XKpMCd]

the idea is that U is always a subset, never a superset, so if so it should not be a problem

[aaronbeall]

So <U within keyof T> would be the same as <U extends keyof T>?

[zpdDG4gta8XKpMCd]

i always forget that subtype of a union is a subset of that union, conversely a supertype of a union must be a superset, so you right that within would look like a poor choice of word to indicate that something is a superset of something else, how about A unties B or A relaxes B or loosens, frees, eases etc: https://www.powerthesaurus.org/loosen/synonyms

[zpdDG4gta8XKpMCd]

@aaronbeall problem with Partial<T> for making a supertype of a product type, is that it only works at the top level, so it doesn't really work for my use case, consider:

type Super<T> = Partial<T>;
type Data = { nested: { value: number; }; }
const one: Super<Data> = {}; // works
const another: Super<Data> = { nested: {} }; // bummer

so i am back to hammering the expected values with the type assertions

expect(data).toEqual(<Data>{ nested: {} });

[aaronbeall]

@Aleksey-Bykov Probably doesn't make this feature any less valuable but for your case I think you can use a recursive partial type:

type DeepPartial<T> = {
  [P in keyof T]?: DeepPartial<T[P]>;
}

(This was suggested as an addition to the standard type defs, which I think would be very useful.)