ES6 destructuring function parameter - naming root object

Solution 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); 
};

Solution 2:

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);
}

Solution 3:

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

1. 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); 
   };
   

2. 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]); 
   };
   

3. 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.