Destructing optional function parameters

user_5471957 0
Viewed 3996

Scratching my head trying to get this optimal. Say I have a function that takes an object as a parameter and destruct it like so:

myfunc = (options = {a:true, b:true, c:true}) => {...}

By default a b and c are true. But say I call myfunc and want b to be false:

myfunc({b:false})

well now options.b === false, but the values for a and c are gone. Is there away I can accomplish this without having to pass in a copy of the default values?

I tried something weird like

myfunc = (options = Object.assign({a:true, b:true, c:true}, options)) =>{}

but that's certainly not right.

destructuring ecmascript-6 javascript
2 Answers

You could use a default object with a destructuring for the wanted properties with the default value without using option. Then create a new one by using short hand properties.

For other objects, you could use, with newer or babeljs, rest properties for objects.

var myfunc = ({ a = true, b = true, c = true, ...rest } = {}) => {
        const option = { a, b, c, ...rest };
        console.log(option);
    };

myfunc();             // take {} as default and the default values inside of the object
myfunc({ b: false }); // take b and default values
myfunc({ foo: 42 });  // take foo and all default values
.as-console-wrapper { max-height: 100% !important; top: 0; }

user_14476750

You can pass default values inside of destructuring statement:

myFunc = ({a=true, b=true, c=true})

Downside obviously is inability to refer to the options as a whole. So maybe having options as a parameter and making destructuring in function body is better option:

myfunc = (options={}) => {
    const {a=true, b=true, c=true} = options;
    ....
}
user_20716970
Related