Extended Error Class - TypeScript

Question

I tried to throw a custom error and print my "CustomError" class name in the console instead of "Error", but without success: classCustomErrorextendsError{constructor(message:string){super(`Lorem "${message}" ipsumdolor.`);this.name='CustomError';}}thrownewCustom

P粉659516906 · Answer

The problem is that when you call super and the new object does not have the expected prototype chain, i.e. it is an instance of Error, not an instance of CustomError .

This problem can be solved elegantly using 'new.target', which is supported since Typescript 2.2, see here: https://www.typescriptlang.org/docs/handbook/release-notes/ typescript-2-2.html

class CustomError extends Error {
  constructor(message?: string) {
    // 'Error' breaks prototype chain here
    super(message); 

    // restore prototype chain   
    const actualProto = new.target.prototype;

    if (Object.setPrototypeOf) { Object.setPrototypeOf(this, actualProto); } 
    else { this.__proto__ = actualProto; } 
  }
}

The advantage of using new.target is that you don't have to hardcode the prototype, like some of the other answers asked here. This has the further advantage that classes inheriting from CustomError will also automatically get the correct prototype chain.

If you were to hardcode the prototype (e.g. Object.setPrototype(this, CustomError.prototype)), CustomError itself would have a working prototype chain, but any class from CustomError inheritance will be broken, e.g. instances of class VeryCustomError will not be the expected instanceof VeryCustomError, but just instanceof CustomError .