ES6 const & let, template literals, destructuring, and arrow functions

Question 1

What is a code block? What are some examples of a code block?

Answer 1

A code block is a chunk of code inside curly braces

if (conditions) { //code block here }

Question 2

What does a block scope mean?

Answer 2

Code in a code block can only be referenced by code that is in the code block

Question 3

What is the scope of a variable declared with const or let?

Answer 3

Block scoped as opposed to function scoped

Question 4

What is the difference between let and const

Answer 4

let is mutable

const values cannot be re-assigned, they are immutable

Question 5

Why is it possible to .push() a new value into a const variable that points to an Array?

Answer 5

Variables are objects, objects are mutable

You are updated the index value of an array, but not re-assigning the value of the variable itself

Question 6

How should you decide on which type of declaration to use?

Answer 6

You should use const when the value of the variable isn’t going to be updated

You should use let when the value of the variable is going to be updated

Question 7

What is the syntax for writing a template literal?

Answer 7

let name = String content here

Surround a string with backticks

Question 8

What is string interpolation?

Answer 8

String interpolation allows you to embed variables and expressions in the string.

JavaScript will automatically replace these variables and expressions by their values

${variable}

Question 9

What is destructing, conceptually?

Answer 9

Taking either properties of an object or indexes of an array to separate them to create specific variables

Question 10

What is the syntax for Object destructuring?

Answer 10

const { property, property2, property3 } = objectName

Question 11

What is the syntax for Array destructuring?

Answer 11

const [ variable, variable2, variable3 ] = arrayName

Question 12

How can you tell the difference between destructuring and creating Object/Array literals?

Answer 12

If the brackets are on the left side of the assignment operator, you are destructuring.

If the brackets are on the right side of the assignment operator, you are creating.

Question 13

When an arrow function’s body is left without curly braces, what changes in its functionality?

Answer 13

Without curly braces, you can only have a return on to the right of the arrow.

Question 14

How is the value of this determined within an arrow function?

Answer 14

With an arrow function, it is determined at definition time

Question 15

What is the syntax for defining an arrow function?

Answer 15

() => line of argument

() => { code block }

(para1, para2) => 1 line of argument

(para1, para2) => { code block }

para => 1 line of argument

para => { code block }