Emma Bostian ✨

Posted on Dec 26, 2018

What’s your favorite JS interview question?

#javascript #discuss

What’s your favorite interview question to ask a candidate?

Top comments (41)

Sergey Ivanov • Dec 26 '18

What will output these example? Why?

for (var i=0; i < 10; i++){
    setTimeout(function(){
        console.log(i);
    }, 1000);
}

How to fix it to output numbers from 0 to 9?

Bharath_Bheemireddy • Dec 27 '18

Using let you will get output numbers from 0 to 9.

jakebman • Dec 27 '18

Feel free to run it and prove me wrong, but I believe that it prints 10 ten times, at 1 second intervals.

It's due to binding on the variable I, not the value of I when you create the lambda.

You need to introduce a local variable in the loop body to fix it.

Akash Kava • Dec 27 '18

You are wrong, if you use for(let i instead of for(var i it will print 0 to 9 correctly. jsfiddle.net/uh86qx1v/1/

Hiatt • Dec 27 '18

Can you explain why that happens? I'm new to ES6 and learned that let has block scope. When used in the for loop, in conjunction with setTimeOut set to 1000, you would think the console.log would run every 1000 ms for each console.log. But this doesn't happen. When I tried it, the sequence of console.logs appear all at once after 1000 ms. Why does that happen? Is it because setTimeOut, when called 10 times, gets pushed to the stack, then after 1000 ms the stack does its thing and all 10 calls are executed simultaneously?

If you take out the setTimeOut and just do console.log within the for (var i loop you get the correct result as well.

jakebman • Dec 27 '18

Yup, you're correct. The callbacks are registered very close to each other - they're all queued up to run at basically the same time.

jakebman • Dec 27 '18

Akash - this is the local variable I was trying to describe: jsfiddle.net/e7gyc5ts/1/ (I needed an IIFE to solve a problem that is better solved by let)

Sergey Ivanov • Dec 27 '18

github.com/ufocoder/javascript.int...

Bharath_Bheemireddy • Dec 28 '18

@hiattzhao youtu.be/Nzokr6Boeaw

Sarath S Nair • Dec 28 '18 • Edited

Using let

for(let i=0; i<10; i++) { setTimeout(()=>console.log(i),i*500); };

Using bind

for(var i=0; i<10;i++) { setTimeout(console.log.bind(null,i), i*100); };

Or you can use IIFE

for(var i=0; i<10; i++) { (function(x){ setTimeout(()=> console.log(x), x*100) })(i); };

Ken Bellows • Dec 28 '18

On the bind example, heads up that in some older browsers you need to pass console as the first argument to any console.<method>.bind(). Not that this is a particularly real-world thing to do, but it's bitten me a few times when trying to debug a promise chain in an older browser; I always love to do promise.then(console.log), but in old browsers this breaks until you do promise.then(console.log.bind(console)).

Pranat • Dec 27 '18

It is an excellent question to test how much you understand JS.

Maciej Bójko • Dec 28 '18 • Edited

Fun trivia: setTimeout can have more than two arguments. After the callback function, and the duration, you can pass args for the callback, as well. Instead of working around setTimeout, use setTimeout.

for (var i = 0; i < 10; i++) {
   setTimeout(console.log, 100*i, i); 
}

(Although I don't even write the for loop, like, ever. It's all map, filter, reduce etc. So I'd rewrite it into something like this:

new Array(10)
   .fill(0)
   .map((x, i) => i)
   .forEach(x => setTimeout(console.log, 100*x, x))

raslanove • Dec 29 '18

for (var i=0; i<10; i++) { setTimeout(function() { console.log(i++ - 10); }, 1000); }

Vaishnav Holkar • Aug 14 • Edited

In for loop replace var with let.As let has block scope .It creates new scope of i for each iteration and each setTimeout callback will have its own i value.
Code -->
Method 1 ->By using let

for (let i=0; i < 10; i++){   
 setTimeout(function(){
        console.log(i);
    }, 1000);
}

Method 2->By using IIFE function

for(var i=0;i<10;i++){
    (function IIFE (j){
    setTimeout(function (){
    console.log(j);
    },j*1000);
    })(i);
}

Method 3 ->

for(var i=0;i<10;i++){
    let j=i;
    setTimeout(function (){
        console.log(j);
    },j*1000);
}

Alexander Plavinski • Dec 26 '18

Why is this your favourite question?

Sergey Ivanov • Dec 27 '18

Because answer on this question will discover what answerer know about:

Scopes
Closures
How event loop works in the browser

Srikanth • Jul 10 '21

for (var i = 0; i < 10; i++) {
setTimeout(
(function (i) {
console.log(i)
})(i),
1000
)
}

Srikanth • Jul 10 '21 • Edited

for (var i = 0; i < 10; i++) {
setTimeout(
(function (i) {
console.log(i)
})(i),
1000
)
}

Antonio Radovcic • Dec 27 '18

For candidates with a clear focus on a specific technology/library/framework/… I always ask "In which case would you not use that?".

Or a similar version of that question.

Like "For which kind of project would Angular be a bad architectural choice?".

Because one size never fits all and it's always a good discussion-starter.

I don't ask weird syntax-questions. It's just rude. It doesn't tell me anything at all about the person. It's exploiting their nervousness for no gain. I want to know if they will make the right calls, and produce working, readable and tested solutions. You know, the stuff that is not one google-search away.

Kyle Stephens • Dec 28 '18 • Edited

Solid approach.

Remember, interviewing is a two-way street. It's just as much about you assessing a prospective employer/boss as it is them assessing you.

If an interviewer tries to "catch you out" with some arbitrary, illegible code or tries to belittle you because you don't know answers to some esoteric problem then it says more about them and their approaches to work than it does about your knowledge/abilities.

Antonio Radovcic • Dec 29 '18

Most def, it's more a conversation-starter, without a "right" answer. If they can argue their claim, perfect.

I think I could argue why React should always be used, sometimes be used, and never be used :-)

Kyle Stephens • Dec 29 '18

Absolutely. Evidence about ways of thinking/approaching a problem are much more revealing about a candidate than their ability to interpret obtuse code like a compiler.

Akash Kava • Dec 27 '18 • Edited

What will be output of

(() => console.log(this)).bind("done")()

var fx = (() => console.log(this));
fx.call("done");

Try in chrome, you will not see "done"

(() => console.log(arguments))("a");

Tiago Romero • Dec 27 '18 • Edited

Great question! I believe this is because arrow functions can't be bound and also don't have arguments, for that we need to use regular function definitions:

(function() { console.log(this) }).bind("done")();

var fx = (function() { console.log(this) });
fx.call("done");

(function() { console.log(arguments) })("a");

Jonas Winzen • Dec 27 '18

You can't bind anything to an arrow functions this. But you can bind arguments to them. ie:

((arg) => console.log(arg)).bind(null, "done")()

would work.

Tiago Romero • Dec 27 '18

Oh nice! Thanks for explaining that.

Alexander Plavinski • Dec 26 '18

I always like to ask which project the candidate is most proud of, like what's their "darling" and why is that so. This always opens room to see what they care for, how can they articulate that and opens room for more questions.

Thorsten Hirsch • Dec 27 '18

Yay, that's also my favourite interview question... and I hope to be asked that question, too, when I'm the candidate.

Ben Halpern • Dec 27 '18

I'd ask what the candidate used for build tooling in personal projects and any previous team projects. I think the answer to this question would tell me a lot about how they approached their work and would lead to an insightful conversation.

Guin White • Jun 13 '20 • Edited

I'm new to JS, and I'm pretty clear on why this logs 10 to the console, but I'm not understanding why it logs it 10 times. Can someone please explain this to a noob?

PS - apologies for reactivating an older thread, but this code actually came up in another context, and when I searched I found this discussion. Thanks for understanding.

vupadhayayx86 • Jun 2 '23

setTimeout(()=>console.log("Hello4"),1004)
setTimeout(()=>console.log("Hello3"),1003)
setTimeout(()=>console.log("Hello2"),1002)
setTimeout(()=>console.log("Hello1"),1001)
setTimeout(()=>console.log("Hello0"),1000)

or
for(let i=0;i<5;i++){
setTimeout(()=>console.log("Hello" +i),1000*1)
}

Both code will produce same result!