class constructor properties are customizable based on the class's generic type

In my base class that utilizes generics, the structure is as follows:

class Base<T extends Record<keyof T, unknown> | undefined = undefined> {

  constructor(a: number, b: () => T) {
    
  }
}

My question is if there is a way to make b optional so that I don't have to explicitly set it to undefined when necessary.

It's easy to achieve this with regular functions. Here's an example:

function f<T extends undefined>(a: number, b?: () => T): void;
function f<T extends Record<keyof T, unknown>>(a: number, b: () => T): void;
function f<T extends Record<keyof T, unknown> | undefined>(a: number, b?: () => T): void {

}

Answer №1

Have you explored the Partial type in TypeScript?

This feature allows you to create a constructor where you can specify only certain properties at a given time, giving you control over object initialization. The constructor method is where you can handle these specifics.

I must admit, I find your Type Definition for Base T a bit complex. However, if you're looking to instantiate objects with specific parameters, using Partial could be the solution.

class Base<T> {

  constructor(init?: Partial<Base<T>) {
    
  }

  prop1: string;
  prop2: string;
}

// To demonstrate, here's how you would use it by setting only one property.

var item = new Base<T>({ prop1: "example" });

If you'd like more information on this topic, check out this informative blog post:

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