Before proceeding, let me explain very briefly about Primitive and Reference types in Javascript.

Primitive and Reference type

Javascript has 5 data types that can contain values. Number, string, Boolean , null and undefined. These are referred to as primitive data types. Basically these are passed by value.
Javascript has 3 data types that are passed by reference. Object,Functions and Arrays. Technically, these are all objects and called as reference data types
Primitive data types are immutable, which means their value cannot be modified once created.
Whereas, objects and arrays are mutable , meaning their value can be altered after creation.

Consider the following examples.

Let us start by copying/cloning a string

let message1 = 'hello everyone'let message2 = message1message2 = 'hello world'console.log(message1) // 'hello everyone' console.log(message2) // 'hello world' 

So for so good,, everything is working as expected .
Let us implement the same logic for object

let myObj1 = {  name : 'John Doe',  age: 34}let myObj2 = myObj1myObj2.name = 'Michael'console.log(myObj1) //{name: 'Michael', age: 34}  -> why does the original object `myObj2` got affected ?console.log(myObj2) //{name: 'Michael', age: 34} 

That's because objects are reference type(passed by reference). So, when you use '=', the pointer to the memory address of myObj1 is copied to myObj2. Not the actual value is copied. Since, reference type don't hold values, they are pointer to the value in memory space.

So,, how do you clone an object?

We can use variety of techniques such as the spread operator(...) or Object.assign() or with JSON.parse() and JSON.stringify().

1. Using spread (...)

let myObj1 = {  name: 'John Doe',  age: 34}let myObj2 = {...myObj1} myObj2.name = 'Michael'console.log(myObj1) // {name: "John Doe", age: 34}  Original object `myObj1` is not changedconsole.log(myObj2) // {name: "Michael", age: 34} 

2. Using Object.assign()

let item1 = {  pen: 23,  pencil: 45}let item2 = Object.assign({}, item1)item2.pen = 100console.log(item1) // {pen: 23, pencil: 45}  Original object `item1` is not changedconsole.log(item2) // {pen: 100, pencil: 45} 

Note

Both of the above techniques only does a shallow copy(only the outer part). If you want to perform a deep copy(nested part), there are a few approaches to this problem. The simple technique is by using our well known friend JSON.parse() and JSON.stringify(). But, it is not recommended to do deep copy using this technique, instead , you can use libraries such
as lodash, which is easy to implement and also robust.

3. Using JSON (Not Recommended)

let obj1 = {a:1,b:2,c:{d :3}}let obj2 = JSON.parse(JSON.stringify(obj1))obj2.c.d = 45;console.log(obj1) // {a: 1,b: 2,c: {d: 3}} console.log(obj2) // {a: 1,b: 2,c: {d: 45}} 

Note

If you try to do the above example with spread or Object.assign(),it only modifies
the outer part(shallow).

For example,

let obj = {a:1,b:2,c:{d :3}}let shallowObj = {...obj}shallowObj.b = 20shallowObj.c.d = 30console.log(shallowObj) //{a: 1,b: 20,c: {d: 30}} console.log(obj) //{a: 1,b: 2,c: {d: 30}}   -> Notice that only `b` value is not modified from the original `obj`, whereas, `d` value is modified

A shallow copy means the first level is copied, deeper levels are referenced.
This is where, shallow and deep copy plays an important role.

Thanks for reading my post

Additional Resources

MDN Web Docs:Object.assign()
MDN Web Docs:JSON.parse()
MDN Web Docs:JSON.stringify()
Shallow vs Deep Copy
Primitive vs Reference Type
Why should you use lodash for deep copy