The Object() constructor turns the input into an object. Its behavior depends on the input's type.

Syntax

new Object()
new Object(value)

Object()
Object(value)

Parameters

value Optional

Any value.

Return value

When the Object() constructor itself is called or constructed, its return value is an object:

• If the value is null or undefined, it creates and returns an empty object.
• If the value is an object already, it returns the value.
• Otherwise, it returns an object of a type that corresponds to the given value. For example, passing a BigInt primitive returns a BigInt wrapper object.

When Object() is constructed but new.target is not the Object constructor itself, the behavior is slightly different — it initializes a new object with new.target.prototype as its prototype. Any argument value is ignored. This may happen, for example, when Object() is implicitly called via super() in the constructor of a class that extends Object. In this case, even if you pass a number to super(), the this value inside the constructor does not become a Number instance.

Examples

Creating a new Object

const o = new Object();
o.foo = 42;

console.log(o);
// { foo: 42 }

Using Object given undefined and null types

The following examples store an empty Object object in o:

const o = new Object();

const o = new Object(undefined);

const o = new Object(null);

Obtaining wrapper objects for BigInt and Symbol

The BigInt() and Symbol() constructors throw an error when called with new, to prevent the common mistake of creating a wrapper object instead of the primitive value. The only way to create a wrapper object for these types is to call Object() with them:

const numberObj = new Number(1);
console.log(typeof numberObj); // "object"

const bigintObj = Object(1n);
console.log(typeof bigintObj); // "object"

const symbolObj = Object(Symbol("foo"));
console.log(typeof symbolObj); // "object"

Specifications

Browser compatibility

See also

• Object initializer