How to rewrite a method in JS so that it is called with one more argument by default?

Question

typescript

I have a class, in my case it is a Subject in which I want to override the pipe so that, in addition to everything that is passed to it, it passes catchError to itself. There is an example on the stack https://stackblitz.com/edit/typescript-mvqpgt

Here we have an ExampleClass that has a showParams method, and I need this method to work differently and I want to change its behavior after class initialization (object creation ). I need that in addition to the arguments that this method will take, it will be called with one more argument at the end.

Answer 1

1. Are you sure that you want to patch the whole class and all its descendants, and not its specific instance?
2. You have a number of errors not related to ts in any way, this line: contains two errors: a) calling inside you teach infinite recursion, you must save the old one into a separate variable and call through it; b.) takes an array as the second argument, and you parse it in vain here. If the answer to question 1 is "no", then something like this should be:this.showParams.apply(this, ...args);

class ExampleClass {
  showParams(...args) {
    console.log('I have to be calleb');
    console.log(...args);
  }
}

const exampleInstance = new ExampleClass();

exampleInstance.showParams('one', 'Two', 'Three', 456);

exampleInstance.showParams = (showParamsOld => function(...args){
  args.push('Default Argument');
  return showParamsOld.apply(this, args); 
})(exampleInstance.showParams);

exampleInstance.showParams('one', 'Two', 'Three', 456);

If the answer is yes, then correct yourself.
Just remember that monk patching is bad and should only be used as a last resort.