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Nick Scialli (he/him)
Nick Scialli (he/him)

Posted on • Originally published at typeofnan.dev

10 Challenging JavaScript Quiz Questions and Answers

The following questions are intended to be challenging and instructive. If you know exactly how to answer each one, that's great, but if you get some wrong and learn why you got it wrong, I contend that's even better!

Let me know in the comments if you learn anything from the quiz!


If you enjoy this quiz, please give it a 💓, 🦄, or 🔖 and consider:


Question 1: IIFE, HOF, or Both

Does the following snippet illustrate an Immediately-Invoked Function Expression (IIFE), a Higher-Order Function (HOF), both, or neither?

((fn, val) => {
  return fn(val);
})(console.log, 5);
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Answer and Explanation

Answer: Both IIFE and HOF

The snippet clearly illustrates an IIFE as we immediately invoke a function by passing console.log and 5 to it. Additionally, we find that this is a HOF as fn is a function, and a HOF is defined as any function that takes another function as a parameter or returns a function.


Question 2: Array-to-Object Efficiency

Both a and b are objects with the same properties and values. Which is created more efficiently?

const arr = [1, 2, 3];

const a = arr.reduce(
  (acc, el, i) => ({ ...acc, [el]: i }),
  {}
);

const b = {};
for (let i = 0; i < arr.length; i++) {
  b[arr[i]] = i;
}
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Answer and Explanation

Answer: b

When b is being set, the b[arr[i]] property is set to the current index on each iteration. When a is being set, the spread syntax (...) will create a shallow copy of the accumulator object (acc) on each iteration and additionally set the new property. This shallow copy is more expensive than not performing a shallow copy; a requires the construction of 2 intermediate objects before the result is achieved, whereas b does not construct any intermediate objects. Therefore, b is being set more efficiently.


Question 3: Batman v. Superman

Consider the following superheroMaker function. What gets logged when we pass the following two inputs?

const superheroMaker = a => {
  return a instanceof Function ? a() : a;
};

console.log(superheroMaker(() => 'Batman'));
console.log(superheroMaker('Superman'));
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Answer and Explanation

Answer: "Batman" "Superman"

When passing () => 'Batman' to superheroMaker, a is an instance of Function. Therefore, the function gets called, returning the string "Batman". When passing "Superman" to superheroMaker, a is not an instance of Function and therefore the string "Superman" is just returned. Therefore, the output is both "Batman" and "Superman".


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Question 4: Object Keys, Object Values

Consider the following object.

const obj = {
  1: 1,
  2: 2,
  3: 3
};
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Is Object.keys equal to Object.values?

console.log(Object.keys(obj) == Object.values(obj));
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Answer and Explanation

Answer: false

In this case, Object.keys converts the keys to be string ["1", "2", "3"] and Object.values gives [1, 2, 3]. Even if the values turn out to be the same type, the two arrays are both different objects in memory, so the equality comparison will return false. You will see a lot of quiz questions here drilling into the concepts of object and array comparison!


Question 5: Basic Recursion

Consider the following recursive function. If we pass the string "Hello World" to it, what gets logged?

const myFunc = str => {
  if (str.length > 1) {
    return myFunc(str.slice(1));
  }

  return str;
};

console.log(myFunc('Hello world'));
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Answer and Explanation

Answer: "d"

The first time we call the function, str.length is greater than 1 ("Hello World" is 11 characters), so we return the same function called on str.slice(1), which is the string "ello World". We repeat this process until the string is only one character long: the character "d", which gets returned to the initial call of myFunc. We then log that character.


Question 6: Function Equality

What gets logged when we test the following equality scenarios?

const a = c => c;
const b = c => c;

console.log(a == b);
console.log(a(7) === b(7));
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Answer and Explanation

Answer: false true

In the first test, a and b are different objects in memory; it doesn't matter that the parameters and return values in each function definition are identical. Therefore, a is not equal to b. In the second test, a(7) returns the number 7 and b(7) returns the number 7. These primitive types are strictly equal to each other.

In this case, the equality (==) vs identity (===) comparison operators don't matter; no type coercion will affect the result.


Question 7: Object Property Equality

a and b are different objects with the same firstName property. Are these properties strictly equal to each other?

const a = {
  firstName: 'Bill'
};

const b = {
  firstName: 'Bill'
};

console.log(a.firstName === b.firstName);
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Answer and Explanation

Answer: true

The answer is yes, they are. a.firstName is the string value "Bill" and b.firstName is the string value "Bill". Two identical strings are always equal.


Question 8: Function Function Syntax

Let's say myFunc is a function, val1 is a variable, and val2 is a variable. Is the following syntax allowed in JavaScript?

myFunc(val1)(val2);
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Answer and Explanation

Answer: yes

This is a common pattern for a higher-order function. If myFunc(val1) returns a function, then that function will be called with val2 as an argument. Here's an example of this in action that you can try out:

const timesTable = num1 => {
  return num2 => {
    return num1 * num2;
  };
};

console.log(timesTable(4)(5));
// 20
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Question 9: Object Property Mutation

Consider objects a and b below. What gets logged?

const a = { firstName: 'Joe' };
const b = a;
b.firstName = 'Pete';
console.log(a);
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Answer and Explanation

Answer: { firstName: 'Pete' }

When we set b = a in the second line, b and a are pointing to the same object in memory. Changing the firstName property on b will therefore change the firstName property on the only object in memory, so a.firstName will reflect this change.


Question 10: Greatest Number in an Array

Will the following function always return the greatest number in an array?

function greatestNumberInArray(arr) {
  let greatest = 0;
  for (let i = 0; i < arr.length; i++) {
    if (greatest < arr[i]) {
      greatest = arr[i];
    }
  }
  return greatest;
}
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Answer and Explanation

Answer: no

This function will work fine for arrays where at least one value is 0 or greater; however, it will fail if all numbers are below 0. This is because the greatest variable starts at 0 even if 0 is greater than all array elements.


If you enjoy this quiz, please give it a 💓, 🦄, or 🔖 and consider:


Want more quiz questions? Head over to https://quiz.typeofnan.dev for 72 JavaScript quiz questions!

Top comments (18)

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pentacular profile image
pentacular

identity (===)

Just noting that === is not an identity operator -- it's the strict equality comparison operator.

We can easily tell that it is not an identity operator because, e.g., NaN === NaN is false.

b and a are pointing to the same

It's a bit problematic to say this, since Javascript doesn't have pointers.

It's sufficient to say that a and b have the same value, so they are the same object, and so provide access to the same properties. The rest then follows as you've said.

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objectmatrix profile image
Ahm A • Edited

the equality operator (==) will attempt to make the data types the same before proceeding

the identity operator (===) requires both data types to be the same, as a prerequisite.

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pentacular profile image
pentacular

We can tell that === is not an identity operator since it may return false when comparing a value with itself.

e.g. NaN === NaN.

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objectmatrix profile image
Ahm A • Edited

NaN === NaN // false, The identity evaluation algorithm (IEA) rule 4
Operands are the same type (numbers), but the IEA rule 4 indicates than nothing is equal with a NaN.
The result is false.

The identity evaluation algorithm (IEA) ===:

1. If both operands have different types, they are not strictly equal
2. If both operands are null, they are strictly equal
3. If both operands are undefined, they are strictly equal
4. If one or both operands are NaN, they are not strictly equal
5. If both operands are true or both false, they are strictly equal
6. If both operands are numbers and have the same value, they are strictly equal
7. If both operands are strings and have the same value, they are strictly equal
8. If both operands have reference to the same object or function, they are strictly equal
9. In all other cases operands are not strictly equal.
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Also,NaNs are the only non-reflexive value, i.e., if x !== x, then x is a NaN

more details:
dmitripavlutin.com/the-legend-of-j...

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pentacular profile image
pentacular

The page you've referenced is mostly correct, but they're a bit confused on terminology, and have made things rather more complicated that necessary.

See: ecma-international.org/publication...

7.2.16 Strict Equality Comparison
The comparison x === y, where x and y are values, produces true or false. Such a comparison is performed as follows:

  1. If Type(x) is different from Type(y), return false.
  2. If Type(x) is Number or BigInt, then a. Return ! Type(x)::equal(x, y).
    1. Return ! SameValueNonNumeric(x, y).

Now, what this means is that some things with the same identity compare false with ===, and some things with different identities compare true with ===.

e.g.

NaN === NaN is false

-0 === 0 is true

Which means that it is not checking the identity of the operands -- it is computing a kind of equality (using the strict equality comparison).

So, === is not an identity operator, and for these reasons is not called an identity operator in the language specification.

It's just some confused people on the internet who are making that into a popular mistake. :)

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thgh profile image
Thomas Ghysels

Some instructive feedback for question 10: initialize greatest to -Infinity (or arr[0])

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cayhorstmann profile image
cayhorstmann

These are very nice questions that probe JavaScript fundamentals and not ephemeral trivia. I suggest you change #4 and #6 so that they don't require knowledge about the behavior of == vs. ===. (Nobody should use == with JavaScript in 2020.) If you think that makes #4 too easy, perhaps you could add whether Object.keys(obj).map(x => parseInt(x)) === Object.values(obj). And as a bonus puzzler, why isn't Object.keys(obj).map(parseInt) the same as Object.keys(obj).map(x => parseInt(x))?

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voiedev profile image
Charles Allen

Love the bonus question! That's quite the trap.

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pris_stratton profile image
pris stratton

Very cool I love these. I am going to try the quiz in the link =)

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dotnetcarpenter profile image
Jon Ege Ronnenberg • Edited

Nice quiz. I enjoyed it more than other JS quizzes I have seen.
In Question 2 though, since it is about performance, it might be useful to know that a var declaration in a for loop is more performant than a let declaration. As long as your for loop is scoped to a function, you should not get any nasty surprises.

An example could be:

function each (f, array) {
  for (var i = 0, max = array.length; i < max; ++i) {
    f(array[i], i);
  }
}
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madza profile image
Madza
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naresh profile image
Naresh Poonia

I'm learning Javascript, I have bookmarked this post.
I would like to come here later and check how many I can answer.
Thank you for this post

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alin11 profile image
Ali Nazari
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yogeshdharya profile image
YogeshDharya

Hey Nick ! I really liked the blog and wanted to tell you that as of now(16th March 2023 22:02 IST) my Like made your existing likes sum up to 200 . Than u so much for your work . Kindly keep Sharing your knowledge with us

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alin11 profile image
Ali Nazari

Nice. Learnt many things. Thanks!

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Emrah Korkmaz

Cool post Nick, I've subscribed to your newsletter to getting some quality posts 🤟