Destructure object parameter, but also have refere

Question

This question already has answers here:

ES6 destructuring function parameter - naming root object (3 answers)

Closed 2 years ago.

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)
}

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.

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)
}