In JS we trust - The best way to learn is by building/coding and teaching. I create the challenges to help my friends learn JavaScript and in return it helps me embrace the language in much deeper level. Feel free to clone, fork and pull.
function a(x) {
x++;
return function () {
console.log(++x);
}
}
a(1)();
a(1)();
a(1)();
let x = a(1);
x();
x();
x();- A:
1, 2, 3and1, 2, 3 - B:
3, 3, 3and3, 4, 5 - C:
3, 3, 3and1, 2, 3 - D:
1, 2, 3and3, 3, 3
Answer
This question reminds us about Closure in JS. Closure allows us to create a stateful function and such function can access to variable outside of its scope. In a nutshell, a closure can have access to global variable (scope), father function scope and its own scope.
We have here 3, 3, 3 and 3, 4, 5 because first we simply call the function a(). It works like a normal function and we do not see something stateful here. In later case, we declare a variable x and it stores the value of function a(1), that is why we get 3. 4. 5 rather than 3, 3, 3.
This kind of gotcha gives me the feeling of static variable in PHP world.
function Name(a, b) {
this.a = a;
this.b = b;
}
const me = Name("Vuong", "Nguyen");
console.log(!(a.length - window.a.length));- A:
undefined - B:
NaN - C:
true - D:
false
Answer
We get true in the console. The tricky part is when we create an object from the constructor function Name but we DO NOT USE new keywork. That makes the variable a global one and get the value "Vuong". Remember that it is actually a property of the global object window (in the browser) or global in the nodejs.
We then get a.length ~ 5 and window.a.length ~ 5 which return 0. !0 returns true.
Imagine what would happen when we create the instance me with the new keywork. That is an interesting inquire!
const x = function (...x) {
let k = (typeof x).length;
let y = () => "freetut".length;
let z = {y: y};
return k - z.y();
};
console.log(Boolean(x()));- A:
true - B: 1
- C: -1
- D:
false
Answer
The spread operator ...x might help us obtain the parameter in the function in the form of array. Yet, in Javascript the typeof array return "object" rather than "array". It is totally odd if you are coming from PHP.
That is said, we now have the length of the string object which returns 6. z.y() simply returns the length of the string 'freetut' (7).
Be aware that the function x() (in the form of function express or anonymous function (if you are coming from PHP) return -1 when being called and when converted to bool with Boolean(-1) return true instead of false. Noted that Boolean(0) return false.
(function js(x) {
const y = (j) => j * x;
console.log(y(s()));
function s() {
return j();
}
function j() {
return x ** x;
}
})(3);- A:
undefined - B: 18
- C: 81
- D: 12
Answer
The function js() can be automatically executed without calling it and known as IIFE (Immediately Invoked Function Expression). Noted the parameter x of the function js is actuallly passed with the value 3.
The value return of the function is y(s())), meaning calling three other functions y(), s() and j() because the function s() returns j().
j() returns 3^3 = 27 so that s() returns 27.
y(s()) means y(27) which returns 27*3 = 81.
Note that we can call declare function BEFORE the function is actually declared but not with expression function.
var tip = 100;
(function () {
console.log("I have $" + husband());
function wife() {
return tip * 2;
}
function husband() {
return wife() / 2;
}
var tip = 10;
})();- A: "I have $10";
- B: "I have $100";
- C: "I have $50";
- D: "I have $NaN";
Answer
We have here an IIFE (Immediately Invoked Function Expression). It means we do not have to call it but it will be excuted automatically when declared. The flow is as: husband() returns wife()/2 and wife() returns tip*2.
We might think that tip = 100 because it is a global variable when declaring with var keyword. However, it is actually undefined because we also have var tip = 10 INSIDE the function. As the variable tip is hoisted with default value undefined, the final result would be D. We know that undefined returns NaN when we try to divide to 2 or multiple with 2.
If we do not re-declare var tip = 10; at the end of the function, we will definately get D.
JS is fun, right?
const js = { language: "loosely type", label: "difficult" };
const edu = {...js, level: "PhD"};
const newbie = edu;
delete edu.language;
console.log(Object.keys(newbie).length);- A: 2;
- B: 3;
- C: 4;
- D: 5;
Answer
This challenge revises the ES6's feature regarding spread operator ... Spread operator is quite useful for retrieving parameter in function, to unite or combine object and array in JavaScript. PHP also has this feature.
In the variable edu, we use ...js (spread operator here) to combine both objects into one. It works in the same way with array.
Then we declare another variable named newbie. IMPORTANT note: By declaring the variable like that, both variables point to the SAME POSITION in the memory. We may have known something like $a = &$b in PHP, which let both varibles work in the same way. We might have known about pass by reference in the case.
Then we have 2 as edu.language is deleted. Both objects now have only two elements.
Now is time to think about coping an object in JS either shallow or deep one.
var candidate = {
name: 'Vuong',
age: 30
};
var job = {
frontend: 'Vuejs or Reactjs',
backend: 'PHP and Laravel',
city: 'Auckland'
};
class Combine {
static get() {
return Object.assign(candidate, job);
}
static count() {
return Object.keys(this.get()).length;
}
}
console.log(Combine.count());- A: 5;
- B: 6;
- C: 7;
- D: 8;
Answer
The buit-in method Object.assign(candidate, job) merges the two objects candidate and job into one object. Then the method Object.keys counts the number of key in the object.
Note that two methods get() and count() are defined as static, so they need to be called statically using Class.staticmethod() syntax. Then the final object get 5 elements.
var x = 1;
(() => {x += 1; ++x})();
((y) => {x +=y; x = x%y;})(2);
(() => x += x)();
(() => x *= x)();
console.log(x);- A: 4;
- B: 50;
- C: 2;
- D: 10;
Answer
Initially x is declared with the value 1. In the first IIFE function, there are two operations. First x becomes 2 and then 3.
In the second IIFE function, x = x + y then the current value is 5. In the second operation, it returns only 1 as it undergoes 5%2.
In the third and fouth IIFE functions, we get 2 x = x + x and then 4 x = x * x. It is more than simple.
$var = 10;
$f = function($let) use ($var) {
return ++$let + $var;
};
$var = 15;
echo $f(10);var x = 10;
const f = (l) => ++l + x;
x = 15;
console.log(f(10));- A: 26 and 26;
- B: 21 and 21;
- C: 21 and 26;
- D: 26 and 21;
Answer
This question illustrates the diffences between PHP and JavaScript when handling closure. In the first snippet, we declare a closure with the keyword use. Closure in PHP is simply an anonymous function and the data is passed to the function using the keyword use. Otherwise, it is called as lambda when we do not use the keyword use. You can check the result of the snippet here https://3v4l.org/PSeMY. PHP closure only accepts the value of the variable BEFORE the closure is defined, no matter where it is called. As such, $var is 10 rather than 15.
On the contrary, JavaScript treats the variable a bit different when it is passed to anonymous function. We do not have to use the keyword use here to pass variable to the closure. The variable x in the second snippet is updated before the closure is called, then we get 26.
Note that in PHP 7.4, we have arrow function and we then do not have to use the keyword use to pass the variable to function. Another way to call a global ariable inside a function in PHP is to use the keyword global or employ the built-in GLOBAL variable $GLOBALS.
let x = {};
let y = {};
let z = x;
console.log(x == y);
console.log(x === y);
console.log(x == z);
console.log(x === z);- A: true true true true;
- B: false false false false;
- C: true true false false;
- D: false false true true;
Answer
Technically, x and y have the same value. Both are empty objects. However, we do not use the value to compare objects.
z is x are two objects referring to the same memory position. In JavaScript, array and object are passed by reference. x and z therefore return true when being compared.
console.log("hello");
setTimeout(()=>console.log("hey"), 1);
setTimeout(()=>console.log("kiora"), 2);
setTimeout(()=>console.log("world"), 0);
console.log("hi");- A: "hello" "hey" "kiora" "world" "hi"
- B: "hello" "hi" "hey" "kiora" "world"
- C: "hello" "hi" "world" "hey" "kiora"
- D: "hello" "hi" "hey" "world" "kiora"
Answer
Given that three setTimeout() functions will be kept in the task queue before jumping back to stack, "hello" and "hi" will be printed first, then A is totally incorrect.
We might have the feeling that three setTimeout() functions should be executed in the order "world" -> "hey" -> "kiora" providing that the time we have set are 0 mil second -> 1 mil second -> 2 mil second respectively. Yet, there is no different between 0 and 1 mil second. That is why we will see "hey" in the next. "world" is being executed then and following by the last on "kiora".
For reference, read this https://stackoverflow.com/questions/8341803/difference-between-settimeoutfn-0-and-settimeoutfn-1
String.prototype.lengthy = () => {
console.log("hello");
};
let x = {name: "Vuong"};
delete x;
x.name.lengthy();- A: "Vuong";
- B: "hello";
- C: "undefined"
- D: "ReferenceError"
Answer
String.prototype.someThing = function () {} is the common way to define a new built-in method for String. We can do the same thing with Array, Object or FunctionName where FunctionName is the function designed by ourself.
That is not challenging to realise that "string".lengthy() always returns hello. Yet, the tricky part lies in the delete object where we might think that this expression will entirely delete the object. That is not the case as delete is used to delete the property of the object only. It does not delete the object. Then we get hello rather than ReferenceError.
Note that if we declare object without let, const or var, we then have a global object. delete objectName then return true. Otherwise, it always returns false.
let x = {};
x.__proto__.hi = 10;
Object.prototype.hi = ++x.hi;
console.log(x.hi + Object.keys(x).length);- A: 10
- B: 11
- C: 12
- D: NaN
Answer
First we have an empty object x, then we add another property hi for x with x.__proto__.hi. Note this is equivalent to Object.prototype.hi = 10 and we are adding to the father object Object the property hi. It means every single object will inherit this propety. The property hi becomes a shared one. Say now we declare a new object such as let y = {}, y now has a propery hi inherited from the father Object. Put it simply x.__proto__ === Object.prototype returns true.
Then we overwrite the property hi with a new value 11. Last we have 11 + 1 = 12. x has one property and x.hi returns 11.
const array = (a) => {
let length = a.length;
delete a[length - 1];
return a.length;
};
console.log(array([1, 2, 3, 4]));
const object = (obj) => {
let key = Object.keys(obj);
let length = key.length;
delete obj[key[length - 1]];
return Object.keys(obj).length;
};
console.log(object({1: 2, 2: 3, 3: 4, 4: 5}));
const setPropNull = (obj) => {
let key = Object.keys(obj);
let length = key.length;
obj[key[length - 1]] = null;
return Object.keys(obj).length;
};
console.log(setPropNull({1: 2, 2: 3, 3: 4, 4: 5}));- A: 333
- B: 444
- C: 434
- D: 343
Answer
This question examines how the delete operator works in JavaScript. In short, it does nothing when we write delete someObject or delete someArray. It nonetheless completely deletes and removes a property of an object when writing something like delete someObject.someProperty. In the case of array, when we write delete someArray[keyNumber], it only removes the value of the index, keep the index intact and the new value is now set to undefined. For that reason, in the code first snippet, we get (the length) 4 elements as in the original array but only 3 properties left in the object passed when the function object() is called, as in the second snippet.
The third snippet gives us 4 as declaring an object's propery to either null or undefined does not completely remove the property. The key is intact. So the length of the object is immutable.
For those who are familiar with PHP, we have unset($someArray[index]) that remove the array element, both key and value. When print_r the array, we might not see the key and value that have been unset. However, when we push (using array_push($someArray, $someValue)) a new element in that array, we might see that the previous key is still kept, but no value and not being displayed. That is something you should be aware of. Have a look at https://3v4l.org/7C3Nf
var a = [1, 2, 3];
var b = [1, 2, 3];
var c = [1, 2, 3];
var d = c;
var e = [1, 2, 3];
var f = e.slice();
console.log(a === b);
console.log(c === d);
console.log(e === f); - A: true true true
- B: false false true
- C: true true false
- D: false true false
Answer
a and b returns false because they point to different memory location even though the values are the same. If you are coming from PHP world, then it will return true obviously when we compare either value or value + type. Check it out: https://3v4l.org/IjaOs.
In JavaScript, value is passed by reference in case of array and object. Hence in the second case, d is the copy of c but they both point to the same memory position. Everything changes in c will result in the change in d. In PHP, we might have $a = &$b;, working in the similar way.
The third one gives us a hint to copy an array in JavaScript using slice() method. Now we have f, which is the copy of e but they point to different memory locations, thus they have different "life". We get false accordingly when they are being compared.
var languages = {
name: ['elixir', 'golang', 'js', 'php', {name: "feature"}],
feature: 'awesome',
};
let flag = languages.hasOwnProperty(Object.values(languages)[0][4].name);
(() => {
if (flag !== false) {
console.log(Object.getOwnPropertyNames(languages)[0].length << Object.keys(languages)[0].length);
} else {
console.log(Object.getOwnPropertyNames(languages)[1].length << Object.keys(languages)[1].length);
}
})();- A: 8
- B: NaN
- C: 64
- D: 12
Answer
The code snippet is quite tricky as it has a couple of different built-in methods handling object in JavaScript. For example, both Object.keys and Object.getOwnPropertyNames are used even thought they are quite similar except that the latter can return non-enumerable properties. You might want to have a look at this thoroughly written reference https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Object/getOwnPropertyNames
Object.values and Object.keys return the property value and property name of the object, respectively. That is nothing new. object.hasOwnProperty('propertyName') returns a boolean confirming whether a property exists or not.
We have flag true because Object.values(languages)[0][4].name returns feature, which is also the name of the property.
Then we have 4 << 4 in the if-else flow that returns the bitwise value, equivalent to 4*2^4 ~ 4*16 ~ 64.
var player = {
name: 'Ronaldo',
age: 34,
getAge: function () {
return ++this.age - this.name.length
}
};
function score(greeting, year) {
console.log(greeting + ' ' + this.name + `! You were born in ${year - this.getAge()}`);
}
window.window.window.score.call(window.window.window.player, 'Kiora', 2019);
score.apply(player, ['Kiora', 2009]);
const helloRonaldo = window.score.bind(window.player, 'Kiora', 2029);
helloRonaldo(); - A: "Kiora Ronaldo! You were born in 1985", "Kiora Ronaldo! You were born in 1985", "Kiora Ronaldo! You were born in 1985"
- B: "Kiora Ronaldo! You were born in 1991", "Kiora Ronaldo! You were born in 1991", "Kiora Ronaldo! You were born in 1999"
- C: "Kiora Ronaldo! You were born in 1991", NaN, "Kiora Ronaldo! You were born in 1980"
- D: "Kiora Ronaldo! You were born in 1991", "Kiora Ronaldo! You were born in 1980", "Kiora Ronaldo! You were born in 1999"
Answer
We can use call(), apply() and bind() to appy a function to any object. At first sight, it seems that three functions do the same thing. Yet there are some situations where they are differently employed to handle respective contexts or solve particular problems.
Of the three, only bind() can be executed after binding. We can create a variable to store the result as helloRonaldo() in the code snippet above. apply() and call() will bind and execute the function at the same time. apply() hints us a ~ array where we need to pass an array as parameter. call() hints us c or comma where we pass parameters with a comma. You might want to have a look at this post https://stackoverflow.com/questions/15455009/javascript-call-apply-vs-bind
Note that window.window.window.score or window.score or simply score do the same thing. It points to the score() function in the global scope.
The correct anwser is D. The score() and getAge() functions are nothing special.
var ronaldo = {age: 34};
var messi = {age: 32};
function score(year, tr, t) {
if (typeof tr === 'function' && typeof t === 'function') {
console.log(`You score ${tr(year, t(this.age))} times`);
}
}
const transform = (x, y) => x - y;
const title = (x) => ++x + x++;
const helloRonaldo = score.bind(ronaldo, 2029, transform, title);
helloRonaldo();
const helloMessi = score.bind(messi, 2029, transform, title);
helloMessi();- A: "You score 1989 times" and "You score 1963 times"
- B: "You score 1959 times" and "You score 1989 times"
- C: "You score 1989 times" and "You score 1953 times"
- D: "You score 1959 times" and "You score 1963 times"
Answer
bind() allows us to bind a function declared with any object. Here we bind score() and both ronaldo and messi.
In score() we pass three parameters year, tr and t in which both tr and t are function. They handle simple things as defined afterwards.
When we bind score() with ronaldo and messi, we pass three parameters as declared in the score() function wherein transform and title are functions.
var person = {};
Object.defineProperties(person, {
'name': {
value: 'Vuong',
enumerable: true
},
'job': {
value: 'developer',
enumerable: true
},
'studying': {
value: "PhD",
enumerable: true
},
'money': {
value: "NZD",
enumerable: false
}
});
class Evaluate {
static checkFlag(obj) {
return Object.getOwnPropertyNames(obj) > Object.keys(obj)
? Object.getOwnPropertyNames(obj)
: Object.keys(obj);
}
}
const flag = Evaluate.checkFlag(person);
console.log(flag.length);- A: 1
- B: 2
- C: 3
- D: 4
Answer
Object.keys(obj) is almost identical to Object.getOwnPropertyNames(obj) except the fact that the latter returns any type of object's property regardless of enumerable. By default enumerable is true when creating object. Using Object.defineProperties or Object.defineProperty we can manually set this option to false.
As such the object person will get 3 usingObject.keys(obj)but 4 with Object.getOwnPropertyNames(obj). In short Object.keys(obj) only returns the property setting the enumerable as true.
const id = 10;
const getID = (...id) => {
id(id);
function id(id) {
console.log(typeof id);
}
};
getID(id);- A: ReferenceError
- B: 10
- C: undefined
- D: 'function'
Answer
When declaring a function inside another function, we are working with Closure in JavaScript. Note that if a function is declared as normal (rather than function expression), it is hoisted. We might see several id in the code snippet above but in fact, some of them does nothing.
The result of the code depending on the operator typeof id, which is function. So id in this operation is the id() function.
var book1 = {
name: 'Name of the rose',
getName: function () {
console.log(this.name);
}
};
var book2 = {
name: {value: "Harry Potter"}
};
var bookCollection = Object.create(book1, book2);
bookCollection.getName();- A: 'Harry Potter'
- B: 'Name of the rose'
- C: ReferenceError
- D: Object object
Answer
Object.create allows us to create an object which is based on another object. If we do not pass the second parameter - book2 in this case - the name property of the object bookCollection will be Name of the rose inherited from the book1. It means we can provide additional properties when declaring object with Object.create.
bookCollection has its own property name and another one inherited from book1. In this case its own property name will show up as it has higher priority. That is why we get 'Harry Potter'.
(() => {
const a = Object.create({});
const b = Object.create(null);
let f1 = a.hasOwnProperty('toString');
let f2 = ('toString' in b);
let result = (f1 === false && f2 === false)?console.log((typeof a.toString()).length):console.log(b.toString());
})();- A: ReferenceError
- B: undefined
- C: 0
- D: 6
Answer
The two objects a and b are created using Object.create() operator. There is a bit of difference between them as a inherits from Object prototype but b is totally empty when we pass the null paramater. Yet hasOwnProperty('toString') always returns false neither a nor b given that toString() is not defined inside these objects. The method however is still available as it is inherited from Object prototype.
Both f1 and f2 return false. Note that we use object.hasOwnProperty('key') and ('key' in object) to check the availability of a key in an object. There is a bit difference between the two as the latter also returns the key inherited. You might want to have a look here: https://stackoverflow.com/questions/455338/how-do-i-check-if-an-object-has-a-key-in-javascript
Then typeof a.toString() returns string, which gives us 6 with the .length property.
If the syntax is odd to you, you might look for 'self-invoking function' and 'arrow function' in JavaScript.
let promise = new Promise((rs, rj)=>{
setTimeout(() => rs(4), 0);
Promise.resolve(console.log(3));
console.log(2);
});
promise
.then(
rs => {
console.log(rs ? rs**rs: rs)
return rs
}
).then(
rs => console.log(rs == 256 ? rs: rs*rs)
)
- A: 3, 2, 256, 256
- B: 3, 2, 256, 16
- C: 256, 16, 3, 2
- D: 16, 256, 3, 2
Answer
We first declare a promise-based code with let and then call it. Given that setTimeout() is an asynchronous action, it will run last even the time is set to 0 in setTimeout(() => rs(4), 0);. Although Promise.resolve(console.log(3)) also returns a promise but it is a Microtasks, then it has a higher priority than Tasks as set by setTimeout(). You might want to have a look at this post https://jakearchibald.com/2015/tasks-microtasks-queues-and-schedules/.
In .then() we chain the result so that we have 4^4 in the first then() and 4*4 in the second then(). Note that return rs returns the original value.
async function f() {
let promise = new Promise((resolve, reject) => {
setTimeout(() => resolve("done!"), 0);
});
setTimeout(()=> console.log("world"), 0);
console.log(await promise);
console.log("hello");
}
f(setTimeout(()=>console.log("kiora"),0));
- A: ReferenceError
- B: done, hello, world
- C: hello, done, world
- D: kiora, done, hello, world
Answer
Though we do not declare any paramater for the function f(), we pass setTimeout(()=>console.log("kiora"),0) when call it. We therefore get 'kiora' first.
Given that the variable promise returns a solved promise and it is called with the keyword await, JavaScript will 'pause' at this line console.log(await promise); till the result is resolved. That is why we get "done" at the next result.
Why we do not get "world" or "hello" at the second ? As JavaScript "pauses" at the line with await keyword, we cannot get "hello" as usual (note that whenever we call setTimeout(), this function will run last because it is an asynchronous task operator), whereas setTimeout(()=> console.log("world"), 0); should always run last.
Here we might see a bit of difference when employing await keyword before asynchronous operator (in this case, we use setTimeout() as an example) or when call the function/operator without it.
function name() {
return new Promise(resolve => {
setTimeout(() => {
resolve('New Zealand');
}, 10);
});
}
function fruit() {
return new Promise(resolve => {
setTimeout(() => {
resolve('Kiwi');
}, 20);
});
}
(async function countryandfruit() {
const getName = await name();
const getFruit = await fruit();
console.log(`Kiora: ${getName} ${getFruit }`);
})();
(async function fruitandcountry() {
const [getName, getFruit] = await Promise.all([name(), fruit()]);
console.log(`Hello: ${ getName } ${ getFruit }`);
})();
- A: Null
- B: Kiora
- C: "Hello: New Zealand Kiwi" -> "Kiora: New Zealand Kiwi"
- D: "Kiora: New Zealand Kiwi" -> "Hello: New Zealand Kiwi"
Answer
Both countryandfruit and fruitandcountry are self invoking functions. Both are declared with the keyword async, it means the code inside will run step by step. It helps us control the flow of data much more concise as compared to Promise-based operator or callback way.
The first function returns "Kiora: New Zealand Kiwi" and the second one ouputs "Hello: New Zealand Kiwi". We might think that the order will be the same but actually the order of the result is reversed because the function with await keyword will run step by step rather than in in parallel as Promise.all. It means fruitandcountry will run faster than countryandfruit.
You might want to have a look at the difference between the two at https://alligator.io/js/async-functions/
class MySort{
constructor(object){
this.object = object;
}
getSort(){
return Object.entries(this.object)[0][1].sort()[Object.values(this.object).length];
}
}
const object = {
month: ["July", "September", "January", "December"]
};
const sortMe = new MySort(object);
console.log(sortMe.getSort())- A: July
- B: September
- C: January
- D: December
Answer
Object.entries returns an array consisting of both key and value from an object while Object.values retuns an array of the values of object and Object.keys gives us an array of keys of the object. As such, Object.entries(object) in the code snippet above gives us a nested array with just one element in which the values are put in another nested array like that [["month", ["July", "September", "January", "December"]]].
For that reason, Object.entries(this.object)[0][1].sort() will actually sort the value array and return a new order as "December" -> "January" -> "July" -> "September". Hence, when we get the element with the index given by [Object.values(this.object).length] we get January because [Object.values(this.object).length] give us 1 (the length of the array given by Object.values);
const flag = ([] !==!!!!! []);
let f = () => {};
console.log((typeof f()).length + (flag.toString().length))- A: NaN
- B: 12
- C: 13
- D: 14
Answer
Comparing two arrays or two objects in JavaScript always return false because both are passed by reference, unlike primitive types such as string, number or boolean. That is why comparing [] and [] using either == or === returns false. The weird part is the !==!!!!! which is equivalent to !==, nothing special. So the flag is true.
In the expression function f(), we use arrow function here but and {} is a part of the function rather than an object. In case you want to return an object, you have to write as let f = () => ({}) or simply using normal way to define function. With the keyword return, we can easily catch the content of the function when using normal way to define function.
Thus, the typeof f() returns undefined rathern object. We then get the length 9 and the flag (true) becomes 'true' (a string, by using toString() function), which returns 3 with the property length. We finally get 13.
(function(a, b, c){
arguments[2] = (typeof arguments).length;
c > 10 ? console.log(c): console.log(++c);
})(1, 2, 3);
- A: 4
- B: 5
- C: 6
- D: 7
Answer
We have a self-invoking function with three parameters declared. Note that arguments inside a function returns an object consisting of the parameters of the function.
The key part here is that when we assign a value to that array (it is array-like, as mentioned above) (or any element), the function will use that value rather than the value from the parameter we pass to it when calling the function. Hence, c will be (typeof arguments).length; (6) rather than 3.
As c has a new value of 6, it is definitely less than 10, so we get the final result console.log(++c), which returns 7.
Note that arguments is not available on arrow functions. See more detailed here https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions/arguments
From ES6 onwards, it is recommended to use ...restParameter given that it is a true array. It means you can manipulate the parameter with native JavaScript functions such as map, reduce or filter.
For PHP developer, we have func_get_args() in PHP that does the same thing, but it will not override the value passed. Check it by yourself at https://3v4l.org/dMfhW
class Calculator{
constructor(a, b){
this.a = a
this.b = b
}
static getFlag(){
return new Array(this.a).length == new Array(this.b).toString().length;
}
getValue(){
return Calculator.getFlag() ? typeof this.a: typeof new Number(this.b);
}
}
const me = new Calculator(5, 5);
console.log(me.getValue());- A: NaN
- B: "string"
- C: "object"
- D: "number"
Answer
We have a class named Calculator. When declaring a new instance of the object, we pass two parameters a and b. These two parameters have the same value but new Array(this.a).length is totally different from new Array(this.b).toString().length because the latter returns a string ",,,," meaning the length 4 while the former returns the length of an array and we therefore get 5.
For that reason getFlag() returns false. In getValue() we get typeof new Number(this.b); which returns object. That is a bit different from typeof b, which returns number.
var name = "Auckland";
const nz = {
name: "Kiwi",
callMe: function(){
return this.name;
}
}
let me = nz.callMe;
let she = nz.callMe.bind(nz);
let result = me() === nz.callMe() ? she(): `${me()} ${she()}`;
console.log(result);
- A: undefined
- B: "Auckland"
- C: "Kiwi"
- D: "Auckland Kiwi"
Answer
The key point in this question involves the keyword this in JavaScript. We have a simple object that contains one method and one string property name.
First, it is important to write down is that let me = nz.callMe; and then call me() is totally different from directly calling nz.callMe(). If we assign a variable to a method delared inside an object, this in that method will behave differently (when we call the variable as a method and when dirrectly call that method). In particular, in the first case, this is the window object while in the second one, this inside the function still points to property name in the object nz. It means me() returns "Auckland" while nz.callMe returns "Kiwi".
Then result will return false and we get the final output value ${me()} ${she()}. Why she() is different from me()? You might easily guess that she still bind to the object nz rather than window object as in me().
const club = {
name: 'Juventus',
player : ['Ronaldo'],
showMePlayer: function() {
this.player.map(function(thename){
console.log(this.name.length);
}, this);
},
showMe: function() {
this.player.forEach(function(thename){
console.log(this.name.length);
}.bind(this));
},
show: function() {
const self = this;
this.player.map(function(thename){
console.log(self.name.length);
});
},
Me: function() {
this.player.map(function(thename){
console.log(this.name.length);
});
},
};
club.showMePlayer();
club.showMe();
club.show();
club.Me();- A: 8 - 8 - 8 - 8
- B: "Juventus" - "Juventus" - "Juventus" - "Juventus"
- C: "Ronaldo" - "Ronaldo" - "Ronaldo" - "Ronaldo"
- D: 8 - 8 - 8 - 0
Answer
The code snippet above is not a big challenge for you I guess. It simply gives you an example of this in different contexts when we declare an anonymous function inside a method of an object. The three first methods are common ways to handle this using this as second parameter in map(), by using bind(this) in forEach (or map()) or by that = thistechnique (we did use seft rathern that).
The last method Me() will cause unexpected result because this.name does not bind to the object club. Note that you might get another result when testing the code on jsbin.com. On Chrome and Firefox, we get 0.
For further information, kindly have a look at http://speakingjs.com/es5/ch17.html#_pitfall_losing_this_when_extracting_a_method
((...a)=>{
const b = ["javascript", "new zealand"];
const c = [...a, typeof a, ...b, "kiwi"];
console.log(c.length + c[0].length);
})(new Array(10));- A: 5
- B: 10
- C: 15
- D: 20
Answer
... can be used in two ways in JavaScript (and PHP) as either spread operator or rest parameter. You might have to check the following article about the two. They are the same as three dots, but the way they are employed vary considerably between the two. https://javascript.info/rest-parameters-spread-operator
We see both spread operator and rest parameter in the code snippet above. First the parameter (...a) in the self-invoking function is of course a rest parameter while the constant c we see the spread operator. In the former case, it simply means that you can pass to the function as many parameter as you want. Note that the typeof a in this case is object even though it is a native array in JavaScript. (I means native array because you might think about array-like if we use arguments. Please have a look at the question 28 or this link https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions/arguments).
Spread operator as in the constant c allows us to combine array. So ...a in the code above is rest parameter when it is used as function parameter but in this case it is the syntax of spread operator.
Finally, we get c with 5 elements but the first element has 10 child elements (when we pass to the function new Array(10)). The length of both then returns 15.
function Kiora(name, ...career) {
this.name = name;
return Array.isArray(career) === true && typeof career === "object"? {} : "";
}
var student = new Kiora("Vuong");
console.log(student.name);- A: "Vuong"
- B: undefined
- C: ErrorReference
- D: false
Answer
We have a function constructor Kiora (written with a capital letter, but that is optional) that can be used to create object, as the student object in the code above. In the function, we have two parameters with the second one is actually a rest parameter. The typeof operator is object but if we check with Array.isArray(array) it also returns true.
For that reason, Array.isArray(career) === true && typeof career === "object" returns true. Hence the return operator finally returns an object {}.
You might be surprised when console.log(student.name); outputs undefined given that the constructor function Kiora returns an object. Otherwise, we might simply get the value name.
class Filter{
constructor(element){
this.element = element;
}
filter(){
return this.type() === "object" ? this.element[0].name: "hello";
}
type(){
return typeof this.element;
}
}
let countries = [{name: "New Zealand", isdeveloped: true},
{name: "Vietnam", isdeveloped: false}]
let x = new Filter(countries);
const filter = countries.filter((item) =>{
return !item.isdeveloped;
})
console.log(x.filter().length + filter[0].name.length)
- A: 15
- B: 16
- C: 17
- D: 18
Answer
Apologize that the code snippet is a bit longer than usual. But actually it is not really challenging as you might think. You can easily get the correct result after spending a little of time to debug.
First we declare a class that has two methods. The first method filter() will returns the first element of the array (of the propterty element) or simply returns hello depending on the type() method. We know that typeof of array will return object so the filter() method return this.element[0].name.
Try to make you feel confused, we then call the built-in filter() method. This native method returns a new array depending on the condition we pass to the call-back function. Note that !item.isdeveloped means false. It means we get Vietnam.
Finally we get New Zealand.length and Vietnam.length, which in total returns 18.
async function abc(){
console.log(8);
await Promise.resolve(2).then(console.log)
console.log(3)
}
setTimeout(()=>{
console.log(1)
}, 0)
abc()
queueMicrotask(()=>{
console.log(0)
})
Promise.resolve(4).then(console.log)
console.log(6)
- A: 6 - 8 - 3 - 0 - 4 - 2 - 1
- B: 8 - 2 - 3 - 0 - 4 - 6 - 1
- C: 6 - 8 - 2 - 0 - 4 - 3 - 1
- D: 8 - 6 - 2 - 0 - 4 - 3 - 1
Answer
D is correct anwser. The order of the asynchronous code's output depends on the MicroTask or MacroTask. MicroTask has a higher priority. Note that the synchronous code always be executed before asynchronous code. So in essense, we have the order as follows:
1) synchronous code
2) microtask code (promise, queueMicrotask)
3) macrotask code (setTimeout, setInterval)
Be awared that in Nodejs environment, we also have process.nextTick(callback) which has the highest priority but we dont have it in this code.
So, first callback in the setTimeout() will be executed last given that this is a MacroTask. That is why we got 1 last.
Second, the function abc() is called next. Then we have 8 printed out in the console first. As the next line of code inside that function is an asynchrnous code with the keyword "await", we then console.log(6) as Promise.resolve(4).then(console.log) is an asynchrnous code. That is why we got 6.
Now is the time for Promise.resolve(2), so we get 2. At this point, you might have some sort of confusion. What will happend if we do not pass the keyword "await" before Promise.resolve(2) ?
As we have await, the code will be blocked here. Then what? We get 0 and 4 not 3. Promise and queueMicrotask are both microtask and they are already to run before console.log(3). The reason is that microtask queue need to be emptied before any other codes can be called in the callstack.
In the next step, we get 3 and the last one is 1.
What would happend if we do not have the await keyword? Then the order of the output will be 8 - 3 - 6 - 2 - 0 - 4 -1.
function myAccount(money){
let myMoney = money;
return {
status: function(){
return `You have $ ${myMoney} in your account`;
},
dePoSit: function(amount){
myMoney = myMoney + amount;
},
withDraw: function(amount){
if(amount > myMoney){
return `You cannot withdraw money now`;
}
myMoney = myMoney - amount;
}
}
}
const vuong = myAccount(1000);
vuong.withDraw(500);
vuong.withDraw(200);
vuong.dePoSit(100);
vuong.withDraw(50);
console.log(vuong.status());- A: "You have $ 950 in your account"
- B: "You have $ 1000 in your account"
- C: "You have $ 550 in your account"
- D: "You have $ 350 in your account"
Answer
As the "state" of the data is preserved each time we call dePoSit() or withDraw(), hence we get $350 after all.
Noted that that is a kind of "factory" function with "preload" data. You might think about another object when pass to myAccount(somedata); some other data. That is a really helpful way to create multiple objects from a factory function.
const hoccoban = {
x: "youtube.com/hoccoban".length,
getMe(){
const inner = function(){
console.log(++this.x)
}
inner.bind(this)();
}
};
hoccoban.getMe();- A: 20
- B: 21
- C: 22
- D: 23
Answer
We get 21. First "youtube.com/hoccoban" returns 20 as we are using the property length of the string. Then it is being added one more value in ++this.x. The question here seems trivial but it is actually not. There is a crucial note we should keep in mind is that console.log(++this.x) will not work as x is undefined when it is called outside of the object.
We can solve the problem with this in this case by using arrow function in the inner so that is can become something like const inner = () => {} as the arrow function does not actually have this. It will automatically look around and call the available object when the function is executed.
The second solution is that we can somehow "bypass" the tricky this by using that/this solution. We just need to declare a new variable const that = this inside getMe() and before declaring inner function. That is a quite common practice.
The third solution is to take advantage of call(), bind() and apply() which are native methods of function (yes, function is also an object in JavaScript). In this case, we implement bind(this) to "bind" the function and the object so that this can actually point to the object when the function is executed. Note that bind() cannot be instantlly executed so that we need to add () after we bridge the function and the object. If we replace bind() with call(), then we do not need to pass () as in the above example. So inner.bind(this)(); will become inner.call(this);. They are technically equal. In practice, we tend to create a new variable to get the result from the binding of the function and the object.
function* hocCoBan(){
yield "js.edu.vn";
yield "youtube.com/hoccoban";
yield "Vuong Nguyen";
}
let data = hocCoBan();
console.log((typeof data).length + data.next().value.length)- A: NaN
- B: 10
- C: Error
- D: 15
Answer
First, take a closer look at the function. It has a asterisk (*) next to the keyword "function". We do not have return keyword inside the function itself. What is going on here?
It you have already known about generator, then this code snippet is not a big deal at all. We do not use generator very often, but this native JavaScript feature is the basis for async/await function, which is supported in ES7 that allows us to handle the flow of asynchronous code much easily.
The operator typeof data will return object rather than function, which is the same case with typeof hocCoBan(). Of course, typeof hocCoBan still returns function. But it is actually a normal function. Basically, we get 6 in the operator (typeof data).length.
Then data.next() calls the the built-in method next() which will output the value in the first yield, which is declared in the function. Then we get the length 9 with the string js.edu.vn.
After all, we get 15. Not that understanding generator is quite important if you really want to understand async/await function.