Daniel Earwicker Chief Software Architect
CNL Software Ltd

TypeScript and runtime typing


TYPESCRIPT 2016-09-04

Prompted by this question on Reddit.

I'd want to declare a type that points to class extending another class. Please note, a CLASS not INSTANCE. I've tried something like this:

type EventClass = class extends Event;

type Listener = (data: class extends Event) => void;

and later on:

private handlers: Map<EventClass,Listener[]>;

But unfortunately this syntax does not work. How I can declare a type that points to CLASS extending another CLASS?

You want a runtime value that specifies a type of event, so you can use it as the key in a Map.

In Java you'd use the Class class, in C# you'd use Type. These are runtime type representatives that reflect the compile time types (perfectly in C# thanks to reified generics, imperfectly in Java due to erased generics).

But TypeScript is just JavaScript with static typing added to it, in a way that tries to be of maximum usefulness in real-world JavaScript, as it is used.

It's not an attempt to recreate the environment (including runtime services) of any other language. Unfortunately there is no general runtime representation of a type in standard JavaScript.

The JavaScript typeof operator returns a string, and is only useful for primitives. The TypeScript typeof operator (which can only appear in type positions) has no effect on generated JavaScript, so there is no way to sneak that information into the runtime world.

But all is not lost. The JavaScript instanceof operator checks the constructor of an object's prototype. Assuming the pattern has been followed correctly (which it is if you use the class keyword to declare your types), an object created by new C can be said to be an instanceof class C. Note that C is just a function so it is an object at runtime.

Suppose you have:

class FoodEvent extends Event {
    flavour: string;
}

class SecurityEvent extends Event {
    threatLevel: number;
}

And here's all Listener can do:

type Listener = (data: Event) => void;

Note that saying Event in that context is sufficient to mean anything derived from Event, so you don't need a way to say that. This is standard OO polymorphism. (In TypeScript, which has structural typing, it actually means "anything with a compatible type shape".)

The types FoodEvent and SecurityEvent are also functions at runtime, so they can be stored in a variable:

const blah = FoodEvent; // works

How do we declare the type of such a function? In TypeScript we can write:

type Constructor<T> = {
    new(...args: any[]): T;
};

That is, a function that must be called with the new prefix, takes any number of arguments of whatever type, and returns something compatible with type T.

So now we can define our map of handlers (to shorten the example I've simplified it: a single Listener instead of an array of them.)

const handlers = new Map<Constructor<Event>, Listener>();

And I can then register a Listener:

handlers.set(FoodEvent, data => {
   // handle the food-related event        
});

This isn't perfect because data is just of type Event. It has to be, of course. The key type of the handlers map has to be some general type - this is no different from Java or C#.

But we can create a helpful way to register a handler for a known type:

function setHandler<TEvent extends Event>(
    constructor: Constructor<TEvent>,
    handler: (data: TEvent) => void
) {
    handlers.set(constructor, handler);
}

This is where it becomes meaningful to talk about "a type that extends Event". We want a more specific type than Event so we can use it in the handler, but it must be compatible with Event so it fits with the map.

At first glance you might think this is no improvement because the constructor must be passed both as a runtime value as well as a type parameter. We have to repeat ourselves. Doesn't that leave a type hole?

But no! Type inference comes to the rescue:

setHandler(FoodEvent, food => {
    console.log(`Food flavour is ${food.flavour}`);
});

setHandler(SecurityEvent, security => {
    console.log(`Security threat-level is ${security.threatLevel}`);
});

We just specify the event constructor as a runtime parameter, and that is sufficient for TypeScript to "pull out" the return type of the constructor so we can constrain the handler's data parameter. So food is properly typed as FoodEvent etc.

It's a neat example of how TypeScript works strictly within the limits of the existing JavaScript infrastructure, and finds the stuff that works, and gives it full language support at compile time, instead of cooking up some new incompatible approach. TypeScript is JavaScript, only with elegant and powerful static typing added.

Advanced Note: there's something weird about our setHandler function. The implementation is just:

handlers.set(constructor, handler);

Now, handler is of type (data: TEvent) => void, that is: a function that takes a TEvent. But we're passing it to the set method, which in this case is of type Listener, which is just an alias for (data: Event) => void. Notice anything strange?

Here it is broken into steps:

function myFoodHandler(food: FoodEvent) {
    console.log(`Food flavour is ${food.flavour}`);
}

let whateverHandler: (data: Event) => void;

whateverHandler = myFoodHandler; // This is the strange part!

const evt: Event = // ... get an event...

whateverHandler(evt);

We can pass any old Event when we call through whateverHandler. And yet we're able to assign to it a function that expects to receive a specific derived type of Event! That assignment is breaking the rules. TypeScript allows this deliberately. It's unsound by design, in this particular situation, on the basis that this kind of incompatibility rarely leads to bugs and it would be hard to explain to users if the language was rigidly sound.

It certainly doesn't cause us a problem in this case, because we've wrapped the Map in a type-safe helper. But it is worth bearing in mind that this is one kind of type error that TypeScript prefers not to catch.

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