It is feasible to completely override a class in TypeScript

I have a subclass defined as follows:

customException.ts

/**
 * Custom Error class.
 *
 * @class Error
 * @extends {Error}
 */
class Error {
  /**
   * @type {string}
   * @memberof Error
   */
  message: string;

  /**
   * @type {boolean}
   * @memberof Error
   */
  isCustom: boolean;

  /**
   * @type {number}
   * @memberof Error
   */
  statusCode: number;

  /**
   * Constructor for creating a custom error.
   *
   * @param {string} message
   * @param {number} statusCode
   * @memberof Error
   */
  constructor(message: string, statusCode: number) {
    this.isCustom = true;
    this.message = message;
    this.statusCode = statusCode;
  }
}    

All other classes inherit from this base class as shown in:

import Error from customException

ForbiddenError extends Error
and we use it without any issues:

throw new ForbiddenError

However, why doesn't the base class Error need to be defined like:

class Error extends Error, since there's no "extend" keyword used, how does it still exhibit the behavior of the parent class?

Answer №1

The throw statement is used to throw a custom exception.

When you use the throw statement, the expression specifies the value of the exception that will be thrown. The examples below show how different types of exceptions can be generated:

throw 'Error2'; // throws an exception with a string value
throw 42;       // throws an exception with the value 42
throw true;     // throws an exception with the value true
throw new Error('Required');  // throws an error object with the message 'Required'

It is possible to throw any type of data, and there is no requirement to extend the Error class in order to do so.

The comments within the code are for documentation purposes and do not affect the compiler's behavior.

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