Q
What will be the output of the following code?
console.log(a); let a = 10;

Answer & Solution

Answer: Option D
Solution:
let
declarations are hoisted but not initialized, so accessing a before its declaration results in a ReferenceError.
Related Questions on Average

How can you avoid variable hoisting issues with let?

A). Declare variables at the bottom of the code

B). Use var instead

C). Declare variables at the top of their scope

D). Do not declare variables

In which scope are let variables hoisted?

A). Function scope

B). Block scope

C). Global scope

D). Module scope

Which statement about let and const is correct?

A). Both are block scoped, but only let can be re-assigned

B). Both are block scoped, but only const can be re-assigned

C). Only const is block scoped

D). Both are not block scoped

What is the 'temporal dead zone'?

A). The period during which a variable is declared but not yet initialized

B). The time when the variable is accessible throughout the program

C). The period after variable initialization

D). None of the above

What will be the output of the following code?

let i = 50; { let i = 55; console.log(i); } console.log(i);

A). 50 50

B). 55 50

C). 50 55

D). ReferenceError

What will be the output of the following code?

let d; console.log(d); d = 15;

A). undefined

B). 15

C). null

D). ReferenceError

What is the scope of a variable declared with let inside a loop?

A). Global scope

B). Function scope

C). Block scope

D). Module scope

What will be the output of the following code?

let f = 10; if (true) { console.log(f); let f = 20; }

A). 10

B). 20

C). undefined

D). ReferenceError

What is the difference between let and var in terms of scope?

A). let is function scoped, var is block scoped

B). let is block scoped, var is function scoped

C). Both are function scoped

D). Both are block scoped

What happens when you try to re-declare a let variable in the same scope?

A). It reassigns the value

B). It throws a SyntaxError

C). It re-declares the variable

D). It throws a TypeError