Problem

Sometimes instanceof might not be enough when working on a library that may potentially be installed multiple times in a project. More on that on my blog.

Solution

As described on my blog - using special @type property to indicate final type of an object.

@pallad/type-check is a tool that helps to achieve it in easy way.

Community

Join our discord server

Installation

npm install @pallad/type-check

Usage

Usage with TypeCheck instance in private scope.

import {TypeCheck} from '@pallad/type-check';

// Create TypeCheck instance with globally unique identifier for class

const CHECK = new TypeCheck<SomeLibraryClass>('some-library-class-unique-identifier')

export class SomeLibraryClass {
	constructor() {
		// assign necessary properties to current instance
		CHECK.assign(this);
	}

	static isType = CHECK.isType
}


SomeLibraryClass.isType(new SomeLibraryClass()); // true

Usage with TypeCheck.clazz.

import {TypeCheck} from '@pallad/type-check';

const CHECK = new TypeCheck<SomeLibraryClass>('some-library-class-unique-identifier')

export class SomeLibraryClass extends CHECK.clazz {
	// properties assignment are done in constructor automatically
	// static `isType` method is also assigned
}

SomeLibraryClass.isType(new SomeLibraryClass()); // true

About type identifiers

Type identifiers (provided in TypeCheck's constructor) must be globally unique so there is no potential conflict with other type check from other libraries. Therefore recommended pattern fo identifier is: [packageName]/[className] For example:

• @pallad/type-check/SomeClassName
• @pallad/secret/Secret

Typescript

Obviously everything is properly typed and proper type guards are applied

const CHECK = new TypeCheck<Foo>('foo')

export class Foo extends CHECK.clazz {
	prineMe() {
		return 'hello';
	}
}

if (Foo.isType(value)) {
	// at this stage Typescript knows that `value` is a type of `Foo`
	value.printMe();
}

Multiple types

Object can can hold multiple types for flexibility purposes.

const CHECK1 = new TypeCheck('foo')
const CHECK2 = new TypeCheck('bar')

const value = {};
CHECK1.assign(value);
CHECK2.assign(value);

CHECK1.isType(value); // true
CHECK2.isType(value); // true