- Write code to get array of names from given array:
const users = [
{ name: 'John', age: 25, country: 'United States' },
{ name: 'Anis', age: 31, country: 'Algeria' },
{ name: 'Jemmy', age: 50, country: 'France' },
{ name: 'Packard', age: 70, country: 'Germany' },
];Solution
// Using map function
const names = users.map(({ name }) => name);
// Using forEach function
const names = [];
users.forEach(({ name }) => name);
// Using for-of loop
const names = [];
for (const { name } of users) {
names.push(name);
}
// Using for-loop
const names = [];
for (let index = 0; index < users.length; index++) {
names.push(users[index]);
}- Get an array of online users from the given array:
const users = [
{ username: 'Jhon', isOnline: true },
{ username: 'Alex', isOnline: false },
{ username: 'Kelly', isOnline: false },
{ username: 'Joma', isOnline: true },
{ username: 'Peter', isOnline: false },
];Solution
// Using map filter and map functions
const onlineUsers = users
.filter(({ username, isOnline }) => {
if (isOnline) return username;
})
.map(({ username }) => username);
// Using forEach function
const onlineUsers = [];
users.forEach(({ username, isOnline }) => {
if (isOnline) onlineUsers.push(username);
});
// Using for-of loop
const onlineUsers = [];
for (const { username, isOnline } of users) {
if (isOnline) onlineUsers.push(username);
}
// Using for-loop
const onlineUsers = [];
for (let index = 0; index < users.length; index++) {
if (users[index].isOnline) onlineUsers.push(users[index].username);
}- Sort the given array by age:
const students = [
{ name: 'Jhon', age: 20 },
{ name: 'Alex', age: 22 },
{ name: 'Kelly', age: 21 },
{ name: 'Joma', age: 23 },
{ name: 'Peter', age: 24 },
];Solution
// Using map and sort functions
const sortedAges = students
.map(({ age }) => age)
.sort((prev, next) => prev - next);- What is the output? and why?
let variable;
console.log(typeof variable);Solution
Output: undefined.
Because we defined the variable without assigning it any value, and the default value is undefined.
- What is the output? and why?
let variable = null;
console.log(typeof variable);Solution
Output: object.
JavaScript has 7 primitive types. All primitive types , except null can be tested with typeof operator.
typeof null returns object. If we want to check for null we have to use ===null.
| Type | typeof return |
Wrapper |
|---|---|---|
| Null | object |
N/A |
| Undefined | undefined |
N/A |
| Boolean | boolean |
Boolean |
| Number | number |
Number |
| BigInt | bigint |
BigInt |
| String | string |
String |
| Symbol | symbol |
Symbol |
- What is the output? and why?
console.log(variable);
let variable = 1;Solution
Output: ReferenceError: Cannot access 'variable' before initialization.
Javascript has a default behavior which is moving the declaration to the top of the current scope (script or function). We call this behavior Hoisting.
In JavaScript we can declare a variable after it has been used. Hoisting is ofter considered a feature of var declarations.
However, we can consider the following behaviors as Hoisting:
| Hoisting | Behavior |
|---|---|
| Value | Being able to use a variable's value in its scope before the line it is declared |
| Declaration | Being able to reference a variable in its scope before the line it is declared, without throwing a ReferenceError, but the value is always undefined. |
let allows you to declare variables that are limited to the scope of a block statement, or expression on which it is used, unlike the var keyword, which declares a variable globally, or locally to an entire function regardless of block scope. The other difference between var and let is that the latter can only be accessed after its declaration is reached.
For this reason, let declarations are commonly regarded as non-hoisted.
- What is the output? and why?
console.log(variable);
variable = 1;Solution
Output: ReferenceError: variable is not defined.
Because we access a variable that does not exist in the scope. This exception occurs when there is a non-existent variable referenced somewhere.
In JavaScript we declare variables using the keywords:
varletconst
- Create a
counterfunction which hasincrementandgetValuefunctions.
Solution
function counter() {
let counter = 0;
return {
increment: function (value) {
return counter + value;
},
getValue: function () {
return counter;
},
};
}- Write a function
concatenatewhich concatenatesfirstNameandlastName(concatenate(firstName)(lastName)).
Solution
const concatenate = (firstName) => (lastName) => firstName + ' ' + lastName;- Write a
curryfunction.
Function Currying is a concept of breaking a function with many arguments into many functions with single argument in such a way, that the output is same. In other words, its a technique of simplifying a multi-valued argument function into single-valued argument multi-functions.
Solution
The concept of currying function is named after the Haskell Brooks Curry who developed it.
By definition, Currying is a process of transforming a function with higher arity to a function with lower arity. Curry functions are higher order functions which takes a function as input and returns a function that accepts arguments of the input function and either invokes that function returning its result (if at least arity number of arguments have been provided) or returns a function that accepts the remaining arguments and so on
curry :: ((a,b)=>c) -> a -> b -> c
const curry = (fn) => {
const arity = fn.length;
return function f1(...args) {
if (args.length >= arity) {
return fn(...args);
} else {
return function f2(...moreArgs) {
return f1(...args.concat(moreArgs));
};
}
};
};- Write a function that gets an array and an element and returns an array with this element.
Solution
// Using push method ( not recommended )
const pushElement = (arr, el) => {
arr.push(el);
return arr;
};
// Using spread operator
const pushElement = (arr, el) => [...arr, el];
console.log(pushElement([1, 2, 3], 4));- Write a function that merges two arrays and returns an array.
Solution
// Using push method ( not recommended )
const mergeArrays = (arr1, arr2) => {
arr1.push(...arr2);
return arr1;
};
// Using spread operator
const mergeArrays = (arr1, arr2) => [...arr1, ...arr2];
console.log(mergeArrays([1, 2, 3], [4, 5, 6]));- Remove all duplicates from the given array:
const array = [1, 2, 2, 2, 3, 3, 4, 5, 8, 1, 10, 4, 9, 6];Solution
// Using forEach and includes
const removeDuplicates = (arr) => {
let noDuplicates = [];
arr.forEach((el) => !noDuplicates.includes(el) && noDuplicates.push(el));
return noDuplicates;
};
// Using Set
const removeDuplicates = (arr) => [...new Set(arr)];
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