Q
What keyword is used to declare a block-scoped variable in ES6?

Answer & Solution

Answer: Option B
Solution:
The let keyword allows you to declare variables that are limited to the scope of a block statement.
Related Questions on Average

How do you export a function in ES6 modules?

A). export function myFunc() {}

B). module.exports = myFunc;

C). exports.myFunc = function() {}

D). export myFunc = function() {}

How do you declare a constant variable in ES6?

A). const

B). let

C). var

D). function

How do you check if a value is an array in ES6?

A). Array.isArray(value)

B). value.isArray()

C). value instanceof Array

D). typeof value === 'array'

How can you create a new promise in ES6?

A). new Promise(function)

B). Promise(function)

C). promise(function)

D). New Promise(function)

What is the correct syntax to create a class in ES6?

A). class MyClass {}

B). class = MyClass {}

C). new MyClass = class {}

D). create class MyClass {}

What is the output of [...'hello']?

A). ['hello']

B). ['h', 'e', 'l', 'l', 'o']

C). [104, 101, 108, 108, 111]

D). ['h', 'e', 'll', 'o']

What is the result of [...['a', 'b', 'c']]?

A). ['a', 'b', 'c']

B). [['a', 'b', 'c']]

C). ['abc']

D). Error

How do you create a generator function in ES6?

A). function* gen() {}

B). function gen*() {}

C). function*gen() {}

D). function * gen() {}

What does Promise.resolve(5).then(console.log) print?

A). 5

B). undefined

C). null

D). Error

What does the ... operator do when used in a function parameter?

A). It spreads an array

B). It restates the parameters

C). It indicates a default parameter

D). It represents the rest of the arguments