Does Javascript pass by reference?

Does Javascript pass by reference?

Does Javascript pass by references or pass by values? Here is an example from Javascript: The Good Parts. I am very confused about my parameter for the rectangle function. It is actually undefined, and redefined inside the function. There are no original reference. If I remove it from the function parameter, the inside area function is not able to access it.
Is it a closure? But no function is returned.
var shape = function (config) {
var that = {}; = || “”;
that.area = function () {
return 0;
return that;
var rectangle = function (config, my) {
my = my || {};
my.l = config.length || 1;
my.w = config.width || 1;
var that = shape(config);
that.area = function () {
return my.l * my.w;
return that;
myShape = shape({
name: “Unhnown”
myRec = rectangle({
name: “Rectangle”,
length: 4,
width: 6
console.log( + ” area is ” + myShape.area() + ” ” + + ” area is ” + myRec.area());


Solution 1:

Primitives are passed by value, Objects are passed by “copy of a reference”.

Specifically, when you pass an object (or array) you are (invisibly) passing a reference to that object, and it is possible to modify the contents of that object, but if you attempt to overwrite the reference it will not affect the copy of the reference held by the caller – i.e. the reference itself is passed by value:

function replace(ref) {
    ref = {};           // this code does _not_ affect the object passed

function update(ref) {
    ref.key = 'newvalue';  // this code _does_ affect the _contents_ of the object

var a = { key: 'value' };
replace(a);  // a still has its original value - it's unmodfied
update(a);   // the _contents_ of 'a' are changed

Solution 2:

Think of it like this:

Whenever you create an object in ECMAscript, this object is formed in a mystique ECMAscript universal place where no man will ever be able to get. All you get back is a reference to that object in this mystique place.

var obj = { };

Even obj is only a reference to the object (which is located in that special wonderful place) and hence, you can only pass this reference around. Effectively, any piece of code which accesses obj will modify the object which is far, far away.

Solution 3:

My 2 Cents…. It’s irrelevant whether Javascript passes parameters by reference or value. What really matters is assignment vs mutation.

I wrote a longer, more detailed explanation here (Is JavaScript a pass-by-reference or pass-by-value language?)

When you pass anything (Whether that be an object or a primitive), all javascript does is assign a new variable while inside the function… just like using the equal sign (=)

How that parameter behaves inside the function is exactly the same as it would behave if you just assigned a new variable using the equal sign.. Take these simple examples.

var myString = 'Test string 1';

// Assignment - A link to the same place as myString
var sameString = myString;

// If I change sameString, it will not modify myString, 
// it just re-assigns it to a whole new string
sameString = 'New string';

console.log(myString); // logs 'Test string 1';
console.log(sameString); // logs 'New string';

If I were to pass myString as a parameter to a function, it behaves as if I simply assigned it to a new variable. Now, let’s do the same thing, but with a function instead of a simple assignment

function myFunc(sameString) {

    // Re assignment.. again, it will not modify myString
    sameString = 'New string';

var myString = 'Test string 1';

// This behaves the same as if we said sameString = myString

console.log(myString); // Again, logs 'Test string 1';

The only reason that you can modify objects when you pass them to a function is because you are not reassigning… Instead, objects can be changed or mutated…. Again, it works the same way.

var myObject = { name: 'Joe'; }

// Assignment - We simply link to the same object
var sameObject = myObject;

// This time, we can mutate it. So a change to myObject affects sameObject and visa versa = 'Jack';
console.log(; // Logs 'Jack' = 'Jill';
console.log(; // Logs 'Jill'

// If we re-assign it, the link is lost
sameObject = { name: 'Howard' };
console.log(; // Logs 'Jill'

If I were to pass myObject as a parameter to a function, it behaves as if I simply assigned it to a new variable. Again, the same thing with the exact same behavior but with a function.

function myFunc(sameObject) {

    // We mutate the object, so the myObject gets the change too... just like before. = 'Jill';

    // But, if we re-assign it, the link is lost
    sameObject = { name: 'Howard' };

var myObject = { name: 'Joe'; }

// This behaves the same as if we said sameObject = myObject;
console.log(; // Logs 'Jill'

Every time you pass a variable to a function, you are “Assigning” to whatever the name of the parameter is, just like if you used the equal (=) sign.

Always remember that the equals sign (=) means assignment.
And passing a parameter to a function also means assignment.
They are the same and the 2 variables are connected in exactly the same way.

The only time that modifying a variable affects a different variable is when the underlying object is mutated.

There is no point in making a distinction between objects and primitives, because it works the same exact way as if you didn’t have a function and just used the equal sign to assign to a new variable.

Solution 4:

As with C, ultimately, everything is passed by value. Unlike C, you can’t actually back up and pass the location of a variable, because it doesn’t have pointers just references.

And the references it has are all to objects, not variables. There are several ways of achieving the same result, but they have to be done by hand, not just adding a keyword at either the call or declaration site.

Solution 5:

Function arguments are passed either by-value or by-sharing but never NEVER by reference in Javascript!


Primitive types are passed by-value:

var num = 123, str = "foo";

function f(num, str) {
  num += 1;
  str += "bar";
  console.log("inside of f:", num, str);

f(num, str);
console.log("outside of f:", num, str);

Reassignments inside a function scope are not visible in the surrounding scope.

This also applies to Strings, which are a composite data type and yet immutable:

var str = "foo";

function f(str) {
  str[0] = "b"; // doesn't work, because strings are immutable
  console.log("inside of f:", str);

console.log("outside of f:", str);


Objects, that is to say all types that are not primitives are passed by-sharing. A variable that holds a reference to an object actually holds merely a copy of this reference. If Javascript would pursue a Call-by-reference evaluation strategy, the variable would hold the original reference. This is the crucial difference between by-sharing and by-reference.

What are the practical consequences of this distinction?

var o = {x: "foo"}, p = {y: 123};

function f(o, p) {
  o.x = "bar"; // mutation
  p = {x: 456}; // reassignment
  console.log("o inside of f:", o);
  console.log("p inside of f:", p);

f(o, p);

console.log("o outside of f:", o);
console.log("p outside of f:", p);

Mutating means to modify certain properties of an existing Object. The reference copy that a variable is bound to and that refers to this object remains the same. Mutations are thus visible in the caller’s scope.

Reassigning means to replace the reference copy bound to a variable. Since it is only a copy, other variables holding a copy of the same reference remain unaffected. Reassignments are thus not visible in the caller’s scope like they would be with a Call-by-reference evaluation strategy.

Further information on evaluation strategies in Ecmascript.

Solution 6:

JavaScript is pass by value.
For primitives, primitive’s value is passed.
For Objects, Object’s reference “value” is passed.

Example with Object:

var f1 = function(inputObject){
var f2 = function(){
    var inputObject={"a":1};

calling f2 results in printing out “a” value as 2 instead of 1, as the reference is passed and the “a” value in reference is updated.

Example with primitive:

var f1 = function(a){
var f2 = function(){
    var a =1;

calling f2 results in printing out “a” value as 1.