ES6 destructuring function parameter - naming root object

Question

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}); }; 

Answer 1

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);  };