10 JavaScript Interview Questions

Note: Every question is followed by Output and a rough Explanation. Scroll at your own risk.

Q1.

var msg = "hello";

if (true) {
  var msg = "welcome";
}
console.log(msg);

// ----

let msg2 = "hello";

if (true) {
  let msg2 = "welcome";
}
console.log(msg2);

Output

welcome
hello

Explanation

var is function scoped, therefore, when msg is declared inside the if block, it overrides the msg in the global scope. This does not happen with let as it is block scoped.

Q2.

for (var i = 0; i < 5; i++) {
  setTimeout(function () {
    console.log(i);
  }, 1000);
}

Output

5
5
5
5
5

Explanation

Since var is function scoped the i variable holds the value 5 after the loop ends. The callback function in the setTimeout gets the same value for every occurrence.

Solution

• Convert var to let to create a scope for every iteration.

for (let i = 0; i < 5; i++) {
  setTimeout(function () {
    console.log(i);
  }, 1000);
}

• Wrap the setTimeout in an anonymous function. Passing i as a parameter scopes it to the anonymous function and therefore the value is not lost.

for (var i = 0; i < 5; i++) {
  (function (x) {
    setTimeout(function () {
      console.log(x);
    }, 1000);
  })(i);
}

Q3.

const obj = {
  ["foo_" + (() => 1)()]: "Hello",
};
console.log(obj);

Output

{ foo_1: 'Hello' }

Q4.

console.log([1,2,3] + [4,5,6]);

Output

1,2,34,5,6

Explanation

String([1,2,3]); is "1,2,3"

Hence, "1,2,3" + "4,5,6" is "1,2,34,5,6"

Q5. What's the order of execution?

console.log('1');
setTimeout(() => console.log('2'), 0);
Promise.resolve().then(() => console.log('3'));
console.log('4');

Output

1
4
3
2

Explanation

Priority for event loop is: call stack > microtask > macrotask

All the synchronous code is executed first

Therefore, logs 1, 4

Next, there is a Promise and a setTimeout

Promise callback is stored in microtask queue & setTimeout callback is stored in macrotask queue

microtask has a higher priority over macrotask. Therefore, it logs 3 followed by 2

Q6.

console.log(typeof typeof 1);

Output

string

Explanation

Evaluate from Right to Left

1. typeof 1 return number
2. typeof 'number' returns string

Q7.

console.log(Math.max() < Math.min());

Solution

true

Explanation

The default value for Math.max() is -Infinity & default value for Math.min() is Infinity

Hence, -Infinity < Infinity is true

Q8.

function func() {
  return foo;

  foo = 1;
  function foo() {}
  var foo = "hello";
}

console.log(typeof func());

Output

function

Explanation

Due to one round of parsing (hoisting) the code looks like this

function func() {
  var foo; // hoisted
  function foo() {} // hoisted
  return foo;

  foo = 1;
  foo = "hello";
}

console.log(typeof func());

Because of that, the last value available for foo is function

Q9.

console.log(3 > 2 > 1);

Output

false

Explanation

It starts from left to right
so 3 > 2 equals true

true > 1 is equivalent to 1 > 1 which is false

Q10.

function fn() {
  return this.str;
}

var str = "global";

var obj = {
  str: "obj",
};

console.log(fn(), fn.call(obj));

Output

global obj

Explanation

When executing fn() the value of this is window and window.str is global.

.call() assigns this to obj and obj.str is obj

Note: This solution works in non-strict mode.

---

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Comments

[Mehul Lakhanpal]

In the second example,
The callback functions are asynchronous. Therefore, when the for loop has finished, the variable i has a value 5. And when the callback function starts to execute, at that moment it prints the value which is 5.

[Ibrahim Abdullahi Aliyu]

Hey, can you please explain no 2 more? Why will it be all 5

[Vladimir]

Let's first understand how the loop works.

for (var i = 0; i < 5; i++) {

1. var i = 0 - the first thing that happens
2. i < 5 (0 < 5) is TRUE then we execute the loop
3. i++ (i = 1 now)
4. i < 5 (1 < 5) is TRUE then we execute the loop
5. i++ (i = 2 now)
6. ...
7. i < 5 (4 < 5) is TRUE then we execute the loop
8. i++ (i = 5 now)
9. i < 5 (5 < 5) is FALSE then we stop executing the loop

At the end of the loop i = 5 (not 4). The loop repeated 5 times (i = 0,1,2,3,4).
When setTimeout starts to execute the function it uses the global "i", which is 5, and executes it 5 times hence you see "5" five times.

UPD: replaced "let" by "var". Copied from the wrong place. There are no issues with "let".

[Dheeraj Sharma]

👍👍👍👍👍👍

[Ajithmadhan]

Hy! Can anyone please explain 2 more!?

[Vladimir]

Answered in comments
dev.to/olvnikon/comment/18875

[Mohamed TEBANI]

thank's