ES6 destructuring function parameter - naming root object

Is there a way to retain the name of a destructured function argument? I.e., the name of the root object?

In ES5, I might do this (using inheritance as a metaphor to make the point):

// ES5:
var setupParentClass5 = function(options) {
    textEditor.setup(options.rows, options.columns);
};

var setupChildClass5 = function(options) {
    rangeSlider.setup(options.minVal, options.maxVal);
    setupParentClass5(options); // <= we pass the options object UP
};

I'm using the same options object to hold multiple configuration parameters. Some parameters are used by the parent class, and some are used by the subclass.

Is there a way to do this with destructured function arguments in ES6?

// ES6:
var setupParentClass6 = ({rows, columns}) => {
    textEditor.setup(rows, columns);
};

var setupChildClass6 = ({minVal, maxVal}) => {
    rangeSlider.setup(minVal, maxVal);
    setupParentClass6( /* ??? */ );  // how to pass the root options object?
};

Or do I need to extract all of the options in setupChildClass6() so that they can be individually passed into setupParentClass6()?

// ugh.
var setupChildClass6b = ({minVal, maxVal, rows, columns}) => {
    rangeSlider.setup(minVal, maxVal);
    setupParentClass6({rows, columns});
};

Answers:

I have the 'options' arguments on too many places myself. I would opt for 1 extra line of code. Not worthy in this example, but a good solution when having destructuring on more lines.

const setupChildClass6 = options => {
    const {minVal, maxVal} = options;
    rangeSlider.setup(minVal, maxVal);
    setupParentClass6(options); 
};

You cannot do it directly in the arguments, but you can extract the values afterward:

function myFun(allVals) {
    const { val1, val2, val3 } = allVals;
    console.log("easy access to individual entries:", val2);
    console.log("easy access to all entries:", allVals);
}

You cannot use destructuring and simple named positional argument for the same parameter at the same time. What you can do:

Use destructuring for setupParentClass6 function, but old ES6 approach for setupChildClass6 (I think this is the best choice, just make name shorter):
```
var setupChildClass6 = (o) => {
  rangeSlider.setup(o.minVal, o.maxVal);
  setupParentClass6(o); 
};
```

Use old arguments object. But arguments can slow down a function (V8 particular), so I think it's a bad approach:

var setupChildClass6 = ({minVal, maxVal}) => {
  rangeSlider.setup(minVal, maxVal);
  setupParentClass6(arguments[0]); 
};

ES7 has proposal for rest/spread properties (if you don't need minVal and maxVal in setupParentCalss6 function):
```
var setupChildClass6b = ({minVal, maxVal, ...rest}) => {
  rangeSlider.setup(minVal, maxVal);
  setupParentClass6(rest);
};
```
Unfortunately it's not ES6.

This is the best I could come up with. Not great, but far better (IMO) than doing the destructing on another line. It gives you a heads-up in a way too that you are going to provide options first and foremost, and also that you are destructuring it in the function...

var setupParentClass6 = options => (({rows, columns}) => {
    textEditor.setup(rows, columns);
})(options);

var setupChildClass6 = options => (({minVal, maxVal}) => {
    rangeSlider.setup(minVal, maxVal);
    setupParentClass6(options);
})(options);

ES6 destructuring function parameter - naming root object

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