All primitive data types in JS are copied by value

All objects (incl. arrays and functions) are copied by reference

Example

		let x = 1;
		let y = x; // y stores the copy of "x" value

		// we change the value in "x", but that did not reflect "y"
		x = 9;
		console.log(`x = ${x}`);
		console.log(`y = ${y}`);
	

Demo with arrays

			let arr1 = [1,2,3];
			let arr2 = arr1; // "arr2" stores a reference to the "arr1"

			// we change a value in "arr1", and that WILL reflect the "arr2"
			arr1[0] = 9;
			console.log(`arr1: ${arr1}`);
			console.log(`arr2: ${arr2}`);

			// arr1: 9,2,3
			// arr2: 9,2,3
		

Demo with objects

			let user1 = {
				name: "Maria",
				age: 33
			}
			let user2 = user1; // "user2" stores a reference to the "user1" object

			// we change a value in "user1", and that WILL reflect the "user2" object:
			user1.name = "Pesho";
			console.log(`user1: ${user1.name}`);
			console.log(`user2: ${user2.name}`);

			// user1: Pesho
			// user2: Pesho
		

Shallow copy Array - spread operator - the easiest way

			let arr1 = [1,2,3];
			let arr2 = [...arr1];

			// we change a value in "arr1", and that WILL NOT reflect the "arr2" copy:
			arr1[0] = 9;
			console.log(`arr1: ${arr1}`);
			console.log(`arr2: ${arr2}`);
		

Shallow copy Array - arr.slice() method

			let arr1 = [1,2,3];
			let arr2 = arr1.slice();

			// we change a value in "arr1", and that WILL NOT reflect the "arr2" copy:
			arr1[0] = 9;
			console.log(`arr1: ${arr1}`);
			console.log(`arr2: ${arr2}`);
		

Shallow copy Object - spread operator - the easiest way

			let user1 = {
				name: "Maria",
				age: 33
			};
			let user2 = {...user1}; // make shallow copy

			// we change a value in "user1", and that WILL NOT reflect the "user2" object:
			user1.name = "Pesho";
			console.log(`user1: ${user1.name}`);
			console.log(`user2: ${user2.name}`);
		

Shallow copy Object - Object.assign() method

			let user1 = {
				name: "Maria",
				age: 33
			}
			let user2 = Object.assign({}, user1); // make shallow copy

			// we change a value in "user1", and that WILL NOT reflect the "user2" object:
			user1.name = "Pesho";
			console.log(`user1: ${user1.name}`);
			console.log(`user2: ${user2.name}`);
		

Reference: Object.assign() @MDN

JSON.parse/stringify

Shallow copies did not preserver the cahnge in inner levels, but deep copies does.

			let user1 = {
				name: "Maria",
				age: 33,
				colors:["red","green"]
			}
			let user2 = JSON.parse(JSON.stringify(user1));

			user1.colors[0] = "black";
			console.log(user2.colors);
		

Overview

Object Copying @wikipedia.org

If both operands are objects, then JavaScript compares internal references which are equal when operands refer to the same object in memory.

The equality and strict equality operators will both return true only if two variables point to the same object in memory.

Example

			let arr1 = [1,2,3];

			let arrRef = arr1;
			let arrCopy = [...arr1];

			console.log(`arr1 === arrRef: ${arr1 === arrRef}`);
			console.log(`arr1 === arrCopy: ${arr1 === arrCopy}`);

			// arr1 === arrRef: true
			// arr1 === arrCopy: false