ES6, ES2015: default values for arguments

Tired of typing if (typeof x === 'undefined') x = defaultValue? Again, ES6 will brings us a elegant solution.

#Default values

The = operator can now help to define a default value to a parameter.

#Function parameters

You can now specify a default value for a parameter of a function. This value will be used if the parameter is not defined, or explicitly set to undefined.

function incr(value, step = 1) {
  return value + step;

incr(41); // 42
incr(41, undefined); // 42
incr(33, 9); // 42

#Dynamic default values

We can specify an expression as a default value. This expression will be evalued during the function call, not during the function declaration.

let defaultWho = "world!";
function hello(who = defaultWho.toUpperCase()) {
  return "Hello " + who;

hello(); // 'Hello WORLD!'
defaultWho = "Anyone?";
hello(); // 'Hello ANYONE?'

#Reusing previous parameters

In the default value expression, you can reuse previous function parameters:

function foo(x = 1, y = x + 1) {
  return x + y;

function bar(x = y + 1, y = 1) {
  return x + y;

foo(); // 1 + (1 + 1) → 3
bar(); // (undefined + 1) + 1 → NaN

#Particular case: TDZ (Temporal Dead Zone)

A temporal dead zone is an area of your program where a variable exists, but is not available yet while it didn't get its value.

The following example is obviously invalid:

function foo(x = x) {
  // throws ReferenceError?

In fact, during the call of the function, x has not been defined yet, so it cannot be used as a default value. This example should trigger an error

However, scope rules will make this example invalid:

const x = 1;
function foo(x = x) {
  // 'x' used here is the parameter itself

After the evaluation of the default values, we are in the scope of the function and in this scope, x corresponds to the parameter (not defined yet) and not the value defined above.

ProTip: do not reuse a variable name 3 times (that might helps to read and understand your code).


Like for the function arguments, assignments using destructuring can have a default value.

const obj = { z: 42 };
const { x = 1, y = x + 1, z, w } = obj;
w; // undefined
x; // 1
y; // 2
z; // 42

As a reminder, the assignation in the previous example would have been written this way using ES5:

var x = obj.x === undefined ? 1 : obj.x;
var y = obj.y === undefined ? x + 1 : obj.y;
var z = obj.z;
var w = obj.w;


There is no more reasons a priori to see an undefined in an ES6 codebase.

About the compatibility (at the time of the writing): only Firefox ≥ 43 support this feature (and just for function...). So you will need Babel or Traceur to be able to use it.

Written by naholyr

Associate in ByteClub, developer, consultant, trainer, around modern JavaScript technologies (ES6, Node, React…). Regular open-source contributor, blog posts author, and speaker. Said shortly: passionate :)

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