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10 JavaScript Quiz Questions and Answers to Sharpen Your Skills

Nick Scialli May 27, 2020🚀🚀 9 minute read

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One way we can challenge ourselves to grow as JavaScript developers is to practice with quiz questions! 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!

Question 1: Array Sort Comparison

Consider the following arrays. What gets logged in various sorting conditions?

const arr1 = ['a', 'b', 'c'];
const arr2 = ['b', 'c', 'a'];

console.log(
  arr1.sort() === arr1,
  arr2.sort() == arr2,
  arr1.sort() === arr2.sort()
);

Answer and Explanation

Answer: true, true, false

There are a couple concepts at play here. First, the array sort method sorts your original array and also returns a reference to that array. This means that when you write arr2.sort(), the arr2 array object is sorted.

It turns out, however, the sort order of the array doesn’t matter when you’re comparing objects. Since arr1.sort() and arr1 point to the same object in memory, the first equality test returns true. This holds true for the second comparison as well: arr2.sort() and arr2 point to the same object in memory.

In the third test, the sort order of arr1.sort() and arr2.sort() are the same; however, they still point to different objects in memory. Therefore, the third test evaluates to false.

Question 2: A Set of Objects

Consider the following Set of objects spread into a new array. What gets logged?

const mySet = new Set([{ a: 1 }, { a: 1 }]);
const result = [...mySet];
console.log(result);

Answer and Explanation

Answer: [{a: 1}, {a: 1}]

While it’s true a Set object will remove duplicates, the two values we create our Set with are references to different objects in memory, despite having identical key-value pairs. This is the same reason { a: 1 } === { a: 1 } is false.

It should be noted if the set was created using an object variable, say obj = { a: 1 }, new Set([ obj, obj ]) would have only one element, since both elements in the array reference the same object in memory.

Question 3: Deep Object Mutability

Consider the following object representing a user, Joe, and his dog, Buttercup. We use Object.freeze to preserve our object and then attempt to mutate Buttercup’s name. What gets logged?

const user = {
  name: 'Joe',
  age: 25,
  pet: {
    type: 'dog',
    name: 'Buttercup',
  },
};

Object.freeze(user);

user.pet.name = 'Daffodil';

console.log(user.pet.name);

Answer and Explanation

Answer: Daffodil

Object.freeze will perform a shallow freeze on an object, but will not protect deep properties from being mutated. In this example, we would not be able to mutate user.age, but we have no problem mutating user.pet.name. If we feel we need to protect an object from being mutated “all the way down,” we could recursively apply Object.freeze or use an existing “deep freeze” library.

Question 4: Prototypal Inheritance

In this question, we have a Dog constructor function. Our dog obviously knows the speak command. What gets logged in the following example when we ask Pogo to speak?

function Dog(name) {
  this.name = name;
  this.speak = function() {
    return 'woof';
  };
}

const dog = new Dog('Pogo');

Dog.prototype.speak = function() {
  return 'arf';
};

console.log(dog.speak());

Answer and Explanation

Answer: woof

Every time we create a new Dog instance, we set the speak property of that instance to be a function returning the string woof. Since this is being set every time we create a new Dog instance, the interpreter never has to look farther up the prototype chain to find a speak property. As a result, the speak method on Dog.prototype.speak never gets used.

Question 5: Promise.all Resolve Order

In this question, we have a timer function that returns a Promise that resolves after a random amount of time. We use Promise.all to resolve an array of timers. What gets logged, or is it random?

const timer = a => {
  return new Promise(res =>
    setTimeout(() => {
      res(a);
    }, Math.random() * 100)
  );
};

const all = Promise.all([timer('first'), timer('second')]).then(data =>
  console.log(data)
);

Answer and Explanation

Answer: ["first", "second"]

The order in which the Promises resolve does not matter to Promise.all. We can reliably count on them to be returned in the same order in which they were provided in the array argument.

Question 6: Reduce Math

Math time! What gets logged?

const arr = [x => x * 1, x => x * 2, x => x * 3, x => x * 4];

console.log(arr.reduce((agg, el) => agg + el(agg), 1));

Answer and Explanation

Answer: 120

With Array#reduce, the initial value of the aggregator (here, named agg) is given in the second argument. In this case, that’s 1. We can then iterate over our functions as follows:

1 + 1 * 1 = 2 (value of aggregator in next iteration)
2 + 2 * 2 = 6 (value of aggregator in next iteration)
6 + 6 * 3 = 24 (value of aggregator in next iteration)
24 + 24 * 4 = 120 (final value)

So, 120 it is!

Question 7: Short-Circuit Notification(s)

Let’s display some notifications to our user! What gets logged in the following snippet?

const notifications = 1;

console.log(
  `You have ${notifications} notification${notifications !== 1 && 's'}`
);

Answer and Explanation

Answer: “You have 1 notificationfalse”

Unfortunately, our short-circuit evaluation will not work as intended here: notifications !== 1 && 's' evaluates to false, meaning we will actually be logging You have 1 notificationfalse. If we want our snippet to work correctly, we could consider the conditional operator: ${notifications === 1 ? '' : 's'}.

Question 8: Spread and Rename

Consider the following array with a single object. What happens when we spread that array and change the firstName property on the 0-index object?

const arr1 = [{ firstName: 'James' }];
const arr2 = [...arr1];
arr2[0].firstName = 'Jonah';

console.log(arr1);

Answer and Explanation

Answer: [{ firstName: "Jonah" }]

Spread creates a shallow copy of the array, meaning the object contained in arr2 is still pointing to the same object in memory that the arr1 object is pointing to. So, changing the firstName property of the object in one array will be reflected by the object in the other array changing as well.

Question 9: Array Method Binding

What gets logged in the following scenario?

const map = ['a', 'b', 'c'].map.bind([1, 2, 3]);
map(el => console.log(el));

Answer and Explanation

Answer: 1 2 3

['a', 'b', 'c'].map, when called, will call Array.prototype.map with a this value of ['a', 'b', 'c']. But, when used as a reference, rather than called, ['a', 'b', 'c'].map is simply a reference to Array.prototype.map.

Function.prototype.bind will bind the this of the function to the first parameter (in this case, that’s [1, 2, 3]), and invoking Array.prototype.map with such a this results in those items being iterated over and logged.

Question 10: Set Uniqueness and Ordering

In the following problem, we use the Set object and spread syntax to create a new array. What gets logged (to consider: Are items forced to be unique? Are they sorted?)

const arr = [...new Set([3, 1, 2, 3, 4])];
console.log(arr.length, arr[2]);

Answer and Explanation

Answer: 4 2

The Set object will force unique elements (duplicate elements already in the set are ignored), but will not change order. The resultant arr array will be [3, 1, 2, 4], meaning arr.length is 4 and arr[2] (the third element of the array) is 2.


Nick Scialli

Nick Scialli is a software engineer at the U.S. Digital Service.

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