Destructure object parameter, but also have reference to the parameter as an object? [duplicate]

Question

With ES6 you can destructure objects in function arguments:

({name, value}) => { console.log(name, value) }

The equivalent ES5 would be:

function(params) { console.log(params.name, params.value) }

But what if I would like a reference both to the params object and the nested properties value and name? This is the closest I got, but the drawback is that it won't work with arrow functions because they do not have access to the arguments object:

function({name, value}) {
  const params = arguments[0]
  console.log(params, name, value)
}

Wingfooted

function(params){const {name,value}=params;console.log(params,name,value)}? – Lekishalela

@ProfessorAllman Yeah that comes very close. There's still an assignment line so I guess my question was: is it possible to do that in the function signature, like function({name,value}=params) – Wingfooted

no, it only makes the core hard to read and confusing specially when you start using arrow functions. no real benefit to it. – Jeffereyjefferies

@DayanMorenoLeon I agree, was just curious – Wingfooted

Answer 1

arguments isn't available in arrow functions, and this affects how function parameters should be consistently treated in ES6.

If original parameter is used, it should be destructured inside function:

(param) => {
  const {name, value} = param;
  // ...
}

If several arguments are expected and some argument negotiation takes place (for example, similarly to arguments.length), rest parameter should be used:

(...args) => {
  const [param] = args;
  // ...
}

Boilerplate variable destructuring has its benefits; it's easier this way to debug param or args at breakpoint even if they aren't currently in use.

Connote

Thanks, I think the answer gives a good summary of idiomatic use vs available options – Wingfooted

Sure, you're welcome. I'm using this style myself as a rule of thumb, because param destructuring looks neat, but debugging was an issue for me numerous times in arrow callbacks. – Connote

Answer 2

What about next:

function({name, value}, params=arguments[0]) {
    console.log(params, name, value)
}

Or:

function(params, {name, value} = params) {
    console.log(params, name, value)
}

Brinkman

This doesn't work using arrow functions. – Hoarse

@Hoarse updated, must work now – Brinkman

I find this very confusing because it seems like the function expects 2 arguments. Furthermore it will work for a single parameter, but fails when you do x({name: 'name', value: 'value'}, {name: 'name2', value: 'value2'}) – Wingfooted