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ES6 Default Parameter Expressions

/ 2 min read

I’ve read about something that I wouldn’t probably think about on a normal day: default parameter expressions in ES6 functions.

In ES5 version, we would probably use the following pattern to create a function with default parameter values:

function add(first, second) {
first = first || 1;
second = second || 2;
// do things
}

However, this has its pitfalls. For example, if one of the arguments is 0, then the function would take an OR value, because 0 is falsey. So you’d need to work with typeof. So it would look like that

function add(first, second) {
first = (typeof first !== "undefined") ? first : 1;
second = (typeof second !== "undefined") ? second : 2;
// do things
}

However, today with ES6 we can simply do this

function add(first = 1, second = 2) {
// do things
}

The interesting discovery that I made for myself was that parameters can reference the previous ones, as in

function add(first = 1, second = first) {
// do things
}

Not the other way round though, because parameters are being created in sequence. So if we’re trying to reference the value of a second parameter, we can’t, because it hasn’t been created yet. Just like we wouldn’t with our let and const.

// Okay
let first = 1;
let second = first;
// Not okay
const first = second;
const second = 2;