Variance over type params

Let’s imagine the following situation

We’re writing software that controls machinery at a snack-making factory. Let’s start with a base class and two subclasses

ts
class Snack {
  protected constructor(
    public readonly petFriendly: boolean) {}
}
 
class Pretzel extends Snack {
  constructor(public readonly salted = true) {
    super(!salted)
  }
}
 
class Cookie extends Snack {
  public readonly petFriendly: false = false
  constructor(
    public readonly chocolateType: 'dark' | 'milk' | 'white') {
    super(false)
  }
}Try

The object oriented inheritance at play makes it pretty easy to understand which of these is a subtype of the other. Cookie is a subtype of Snack, or in other words

All Cookies are also Snacks, but not all Snacks are Cookies

Covariance

Our factory needs to model machines that produce these items. We plan for there to be many types of snacks, so we should build a generalized abstraction for a Producer<T>

ts
interface Producer<T> {
  produce: () => T;
}Try

We start out with two kinds of machines

• snackProducer - which makes Pretzels and Cookiess at random
• cookieProducer - which makes only Cookies

ts
let cookieProducer: Producer<Cookie> = {
  produce: () => new Cookie('dark')
};
 
const COOKIE_TO_PRETZEL_RATIO = 0.5
 
let snackProducer: Producer<Snack> = {
  produce: () => Math.random() > COOKIE_TO_PRETZEL_RATIO
    ? new Cookie("milk")
    : new Pretzel(true)
};Try

Great! Let’s try assignments in both directions of snackProducer and cookieProducer

ts
snackProducer = cookieProducer // ✅
cookieProducer = snackProducer // ❌
Type 'Producer<Snack>' is not assignable to type 'Producer<Cookie>'.
  Property 'chocolateType' is missing in type 'Snack' but required in type 'Cookie'.2322Type 'Producer<Snack>' is not assignable to type 'Producer<Cookie>'.
  Property 'chocolateType' is missing in type 'Snack' but required in type 'Cookie'.Try

Interesting! We can see that if we need a snackProducer, a cookieProducer will certainly meet our need, but if we must have a cookieProducer we can’t be sure that any snackProducer will suffice.

Because both of these arrows flow in the same direction, we would say Producer<T> is covariant on T

TypeScript 5 gives us the ability to state that we intend Producer<T> to be (and remain) covariant on T using the out keyword before the typeParam.

ts
interface Producer<out T> {
  produce: () => T;
}Try

Contravariance

Now we need to model things that package our snacks. Let’s make a Packager<T> interface that describes packagers.

ts
interface Packager<T> {
  package: (item: T) => void;
}Try

Let’s imagine we have two kinds of machines

• cookiePackager - a cheaper machine that only is suitable for packaging cookies
• snackPackager - a more expensive machine that not only packages cookies properly, but it can package pretzels and other snacks too!

ts
let cookiePackager: Packager<Cookie> = {
  package(item: Cookie) {}
};
 
let snackPackager: Packager<Snack> = {
  package(item: Snack) {
    if (item instanceof Cookie ) {
      /* Package cookie */
    } else if (item instanceof Pretzel) {
      /* Package pretzel */
    } else {
      /* Package other snacks? */
    }
  }
};
 
cookiePackager = snackPackager;
snackPackager = cookiePackager
Type 'Packager<Cookie>' is not assignable to type 'Packager<Snack>'.
  Property 'chocolateType' is missing in type 'Snack' but required in type 'Cookie'.2322Type 'Packager<Cookie>' is not assignable to type 'Packager<Snack>'.
  Property 'chocolateType' is missing in type 'Snack' but required in type 'Cookie'.Try

If we need to package a bunch of Cookies, our fancy snackPackager will certainly do the job. However, if we have a mix of Pretzels, Cookies and other Snacks, the cookiePackager machine, which only knows how to handle cookies, will not meet our needs.

Let’s build a table like we did for covariance

Because these arrows flow in opposite directions, we would say Packager<T> is contravariant on T

TypeScript 5 gives us the ability to state that we intend Packager<T> to be (and remain) covariant on T using the in keyword before the typeParam.

ts
interface Packager<in T> {
  package: (item: T) => void;
}Try

Invariance

What happens if we merge these Producer<T> and Packager<T> interfaces together?

ts
interface ProducerPackager<T> {
  package: (item: T) => void;
  produce: () => T;
}Try

These machines have independent features that allow them to produce and package food items.

• cookieProducerPackager - makes only cookies, and packages only cookies
• snackProducerPackager - makes a variety of different snacks, and has the ability to package any snack

ts
let cookieProducerPackager: ProducerPackager<Cookie> = {
  produce() {
    return new Cookie('dark')
  },
  package(arg: Cookie) {}
}
 
let snackProducerPackager: ProducerPackager<Snack> = {
  produce() {
    return Math.random() > 0.5
      ? new Cookie("milk")
      : new Pretzel(true)
  },
  package(item: Snack) {
    if (item instanceof Cookie ) {
      /* Package cookie */
    } else if (item instanceof Pretzel) {
      /* Package pretzel */
    } else {
      /* Package other snacks? */
    }
  }
}
 
snackProducerPackager= cookieProducerPackager
Type 'ProducerPackager<Cookie>' is not assignable to type 'ProducerPackager<Snack>'.
  Types of property 'package' are incompatible.
    Type '(item: Cookie) => void' is not assignable to type '(item: Snack) => void'.
      Types of parameters 'item' and 'item' are incompatible.
        Type 'Snack' is not assignable to type 'Cookie'.2322Type 'ProducerPackager<Cookie>' is not assignable to type 'ProducerPackager<Snack>'.
  Types of property 'package' are incompatible.
    Type '(item: Cookie) => void' is not assignable to type '(item: Snack) => void'.
      Types of parameters 'item' and 'item' are incompatible.
        Type 'Snack' is not assignable to type 'Cookie'.
cookieProducerPackager = snackProducerPackager
Type 'ProducerPackager<Snack>' is not assignable to type 'ProducerPackager<Cookie>'.
  The types returned by 'produce()' are incompatible between these types.
    Type 'Snack' is not assignable to type 'Cookie'.2322Type 'ProducerPackager<Snack>' is not assignable to type 'ProducerPackager<Cookie>'.
  The types returned by 'produce()' are incompatible between these types.
    Type 'Snack' is not assignable to type 'Cookie'.Try

Looks like assignment fails in both directions.

• The first one fails because the package types are not type equivalent
• The second one fails because of produce.

Where this leaves us is that ProducerPackager<T> for T = Snack and T = Cookie are not reusable in either direction — it’s as if these types (ProducerPackager<Cooke> and ProducerPackager<Snack>) are totally unrelated.

Let’s make our table one more time

This means that ProducerPackager<T> is invariant on T. Invariance means neither covariance nor contravariance.

Bivariance

For completeness, let’s explore one more example. Imagine we have two employees who are assigned to quality control.

One employee, represented by cookieQualityCheck is relatively new to the company. They only know how to inspect cookies.

Another employee, represented by snackQualityCheck has been with the company for a long time, and can effectively inspect any food product that the company produces.

ts
function cookieQualityCheck(cookie: Cookie): boolean {
  return Math.random() > 0.1
}
 
function snackQualityCheck(snack: Snack): boolean {
  if (snack instanceof Cookie) return cookieQualityCheck(snack)
  else return Math.random() > 0.16 // pretzel case
}Try

We can see that the snackQualityCheck even calls cookieQualityCheck. It can do everything cookieQualityCheck can do and more.

Our quality control employees go through a process where they check some quantity of food products, and then put them into the appropriate packaging machines we discussed above.

Let’s represent this part of our process as a function which takes a bunch of uncheckedItems and a qualityCheck callback as arguments. This function returns a bunch of inspected food products (with those that didn’t pass inspection removed).

We’ll call this function PrepareFoodPackage<T>

ts
// A function type for preparing a bunch of food items
// for shipment. The function must be passed a callback
// that will be used to check the quality of each item.
type PrepareFoodPackage<T> = (
  uncheckedItems: T[],
  qualityCheck: (arg: T) => boolean
) => T[]Try

Let’s create two of these PrepareFoodPackage functions

• prepareSnacks - Can prepare a bunch of different snacks for shipment
• prepareCookies - Can prepare only a bunch of cookies for shipment

ts
// Prepare a bunch of snacks for shipment
let prepareSnacks: PrepareFoodPackage<Snack> = 
  (uncheckedItems, callback) => uncheckedItems.filter(callback)
 
// Prepare a bunch of cookies for shipment
let prepareCookies: PrepareFoodPackage<Cookie> = 
  (uncheckedItems, callback) => uncheckedItems.filter(callback)Try

Finally, let’s examine type-equivalence in both directions

ts
// NOTE: strictFunctionTypes = false
 
const cookies = [
  new Cookie('dark'),
  new Cookie('milk'),
  new Cookie('white')
]
const snacks = [
  new Pretzel(true),
  new Cookie('milk'),
  new Cookie('white')
]
prepareSnacks (cookies, cookieQualityCheck)
prepareSnacks (snacks,  cookieQualityCheck)
prepareCookies(cookies, snackQualityCheck )Try

In this example, we can see that cookieCallback and snackCallback seem to be interchangeable. This is because, in the code snippet above, we had the strictFunctionTypes option in our tsconfig.json turned off.

Let’s look at what we’d see if we left this option turned on (recommended).

ts
// NOTE: strictFunctionTypes = true
 
prepareSnacks (cookies, cookieQualityCheck)
Argument of type '(cookie: Cookie) => boolean' is not assignable to parameter of type '(arg: Snack) => boolean'.
  Types of parameters 'cookie' and 'arg' are incompatible.
    Property 'chocolateType' is missing in type 'Snack' but required in type 'Cookie'.2345Argument of type '(cookie: Cookie) => boolean' is not assignable to parameter of type '(arg: Snack) => boolean'.
  Types of parameters 'cookie' and 'arg' are incompatible.
    Property 'chocolateType' is missing in type 'Snack' but required in type 'Cookie'.
prepareSnacks (snacks,  cookieQualityCheck)
Argument of type '(cookie: Cookie) => boolean' is not assignable to parameter of type '(arg: Snack) => boolean'.2345Argument of type '(cookie: Cookie) => boolean' is not assignable to parameter of type '(arg: Snack) => boolean'.
prepareCookies(cookies, snackQualityCheck )Try

What variance helpers do for you

There are two reasons to use variance helpers in your code

• If you have recursive types in your project, these hints allow TypeScript to type-check significantly faster. Behinds the scenes, the compiler gets to skip a bunch of work, if it knows that a typeParam is purely in or out.
• It allows you to encode more of your intent, and (where useful) catch any changes to variance in the interface declaration instead of at the places where the interface is used.

Here’s a comparison of the error experiences, with and without the variance helpers.

Here’s our working state for Packager<T> again

ts
interface Packager<T> {
  package: (item: T) => void;
}
 
let snackPackager!: Packager<Snack>
let cookiePackager: Packager<Cookie> = snackPackagerTry

And let’s change Packager<T> so that it becomes invariant on T

ts
interface Packager<T> {
  package: (item: T) => void;
  produce: () => T;
}
 
let snackPackager!: Packager<Snack>
let cookiePackager: Packager<Cookie> = snackPackager
Type 'Packager<Snack>' is not assignable to type 'Packager<Cookie>'.
  The types returned by 'produce()' are incompatible between these types.
    Type 'Snack' is not assignable to type 'Cookie'.2322Type 'Packager<Snack>' is not assignable to type 'Packager<Cookie>'.
  The types returned by 'produce()' are incompatible between these types.
    Type 'Snack' is not assignable to type 'Cookie'.Try

Finally, we’ll add that in keyword

ts
interface Packager<in T> {
Type 'Packager<super-T>' is not assignable to type 'Packager<sub-T>' as implied by variance annotation.
  The types returned by 'produce()' are incompatible between these types.
    Type 'super-T' is not assignable to type 'sub-T'.2636Type 'Packager<super-T>' is not assignable to type 'Packager<sub-T>' as implied by variance annotation.
  The types returned by 'produce()' are incompatible between these types.
    Type 'super-T' is not assignable to type 'sub-T'.
  package: (item: T) => void;
  produce: () => T;
}
 
let snackPackager!: Packager<Snack>
let cookiePackager: Packager<Cookie> = snackPackagerTry

The error is surfaced at Packager<T>’s declaration site, and is articulated in terms of violating a variance constraint, not the resultant type-checking error that arises from the call site which requires covariance in order to compile.