Utilizing the in operator for effective type guarding in Typescript

Question

Utilizing the in operator for effective type guarding in Typescript

Before posting an issue on github, I wanted to get some feedback here first.

I'm currently developing a component entity system for a game using typescript. From what I've seen, component entity systems in javascript and typescript can lack type safety, so I have an idea to address this issue.

My plan is to create an Entity type for each system that relies on multiple components. Each Entity type will define the properties that entities processed by that system should have. These Entity types will then be combined into one unified Entity union type.

export type Entity =
    CameraManagerEntity |
    CollisionManagerEntity |
    HoleManagerEntity |
    ParentManagerEntity | ...

Within a system, I want to be able to assert that certain properties exist on an entity and have typescript infer the correct type based on that assertion. For instance, if the CameraManager exports:

interface CameraManagerEntity {
    foo: boolean,
    bar: number
}

And no other type in the Entity union has a "foo" parameter. My expectation is that checking for "foo" within an entity should allow access to "bar" without triggering a type error:

function processEntity(entity: Entity) {
    if ("foo" in entity) {
        return entity.bar; // <-- Type error.
    }
}

Is my approach flawed? It seems like the compiler should be able to determine the specific Entity type based on context. However, it appears that this concept doesn't align with how typescript currently functions. Are there alternative methods to achieve my goal? Appreciate any insights.

Edit: I am familiar with user-defined type guards, but I'm hoping to avoid manual guard implementation as I believe the compiler already possesses the necessary information.

typescript types

Answer 1

Answer №1

Your desired scenario is akin to an inverted Tagged Union Type concept. In a Tagged Union Type, all members share a common field, and TypeScript discriminates based on the value of that field using 'switch'. However, in your current design, union type members are distinguished by unique fields.

I developed a generic type guard that verifies an 'entity' against a type using property presence:

if (hasKeyOf<CameraManagerEntity>(entity, 'camera')) {
  return entity.camera;
}

Below is the complete example:

interface CameraManagerEntity {
  camera: string;
}

interface CollisionManagerEntity {
  collision: string;
}

type Entity = CameraManagerEntity | CollisionManagerEntity;

// Generic type guard 
function hasKeyOf<T>(entity: any, key: string): entity is T {
  return key in entity;
}

function processEntity(entity: Entity) {
  if (hasKeyOf<CameraManagerEntity>(entity, 'camera')) {
    return entity.camera;
  } else if (hasKeyOf<CollisionManagerEntity>(entity, 'collision')) {
    return entity.collision;
  } 
}

Try it in TypeScript Playground

Alternatively, you may want to consider utilizing Tagged Union Types for your entity system, which could offer greater ease-of-use:

interface CameraManagerEntity {
  kind: 'CameraManager';
  camera: string;
}

interface CollisionManagerEntity {
  kind: 'CollisionManager';
  collision: string;
}

type Entity = CameraManagerEntity | CollisionManagerEntity;

function processEntity(entity: Entity) {
  switch (entity.kind) {
    case ('CameraManager'): return entity.camera;
    case ('CollisionManager'): return entity.collision;
  }
}

Try it in TypeScript Playground

Answer 2

Your desired scenario is akin to an inverted Tagged Union Type concept. In a Tagged Union Type, all members share a common field, and TypeScript discriminates based on the value of that field using 'switch'. However, in your current design, union type members are distinguished by unique fields.

I developed a generic type guard that verifies an 'entity' against a type using property presence:

if (hasKeyOf<CameraManagerEntity>(entity, 'camera')) {
  return entity.camera;
}

Below is the complete example:

interface CameraManagerEntity {
  camera: string;
}

interface CollisionManagerEntity {
  collision: string;
}

type Entity = CameraManagerEntity | CollisionManagerEntity;

// Generic type guard 
function hasKeyOf<T>(entity: any, key: string): entity is T {
  return key in entity;
}

function processEntity(entity: Entity) {
  if (hasKeyOf<CameraManagerEntity>(entity, 'camera')) {
    return entity.camera;
  } else if (hasKeyOf<CollisionManagerEntity>(entity, 'collision')) {
    return entity.collision;
  } 
}

Try it in TypeScript Playground

Alternatively, you may want to consider utilizing Tagged Union Types for your entity system, which could offer greater ease-of-use:

interface CameraManagerEntity {
  kind: 'CameraManager';
  camera: string;
}

interface CollisionManagerEntity {
  kind: 'CollisionManager';
  collision: string;
}

type Entity = CameraManagerEntity | CollisionManagerEntity;

function processEntity(entity: Entity) {
  switch (entity.kind) {
    case ('CameraManager'): return entity.camera;
    case ('CollisionManager'): return entity.collision;
  }
}

Try it in TypeScript Playground

Utilizing the in operator for effective type guarding in Typescript

Answer №1

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