Destructuring optional properties

March 29, 2022

I would like to create a function that takes one optional property called verbose, and if it’s not specified it defaults to true. Here is an example:

function Counter(n=0, {verbose=false}) {
    return function() {
        n++;
        if (verbose) console.log(n);
        return n;
    }
}
let c1 = Counter(0, {verbose: true});
c1(), c1(), c1();

Everything works fine. However, as soon as I call it without the object param, I get an error:

function Counter(n=0, {verbose=false}) {
    return function() {
        n++;
        if (verbose) console.log(n);
        return n;
    }
}
let c1 = Counter(0);
// TypeError: Cannot read property 'verbose' of undefined

Why does this occur, as I thought the whole point of having an object that can be destructured in the argument list is the suggested way to have a bunch of optional arguments that can be called. What am I doing wrong here?

>Solution :

Don’t put the default in the destructuring pattern, put it in a default value that’s assigned to the whole object.

function Counter(n=0, {verbose} = {verbose: false}) {
    console.log(verbose);
    return function() {
        n++;
        if (verbose) console.log(n);
        return n;
    }
}
let c1 = Counter(0);
let c2 = Counter(0, {verbose: true});