Better understanding of JavaScript closure inside loops

Question

I'm dealing with the closure issue right now and I need a little help on the topic: Using this code for my training:

for (var i = 1; i <= 10; i++) {
    setTimeout(function(j) {
        console.log(j);
    }(i), 1000);
}

This solution timeouts for a second and then prints 1-10 in ascending order (as expected). Now, I want it to print j once every second:

1 (second passed...)
2 (second passed...)
3 (second passed...)
etc..

How can I make this happen without changing the formation of the loop?

Thanks in advance

CodingIntrigue · Accepted Answer · 2014-10-26 07:38:27Z

3

setTimeout expects a function to be able to execute. You need to return a function and also, seeing as your setTimeout calls will occur one after another, you'll need to delay them

for (var i = 1; i <= 10; i++) {
    setTimeout(function(j) {
        return function() {
            console.log(j);
        }
    }(i), 1000 * i); // Delay
}

jsFiddle

That's without changing the formation of the loop. Although your best solution is to not use a for loop, but chain setTimeout calls instead:

var i = 0;
function start() {
    setTimeout(function() {
        i++;
        console.log(i);
        // If we haven't reached our limit, start again
        if(i < 10) start();
    }, 1000);
}
start();

jsFiddle

answered Oct 26, 2014 at 7:38

CodingIntrigue

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7 Comments

Misha Shblin Over a year ago

I understand there are better ways to solve this, nevertheless Im trying to understand closures so please bear with me... Ive already tried what you suggested on the first attempt, all it did was to print the result an a whole and then wait for 10 seconds before wrapping up. Any other suggestions?

CodingIntrigue Over a year ago

@MishaShblin Absolutely, so the first way answers your question. The problem isn't really with the closure, which you've created correctly, but the fact that setTimeout will only work if it is passed in a function. As it stands you were calling setTimeout(undefined, 1000) because your anonymous function never returned a value

CodingIntrigue Over a year ago

@MishaShblin Didn't see your edit. Check the jsFiddle provided, it should show the code working

Sacho Over a year ago

Your problem isn't related to closures at all - calling setTimeout() 10 times in a loop simply won't make them execute "one after another".

CodingIntrigue Over a year ago

@MishaShblin Nope. Like I said, your closure is fine - you have correctly encapsulated the i variable and modified it so that it is available at a later stage.. Your usage of setTimeout is not. My answer shows the correct your usage of setTimeout, given your correct usage of a closure

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Misha Shblin · Accepted Answer · 2014-10-26 08:09:55Z

1

Thanks RGraham for the solution:

for (var i = 1; i <= 10; i++)(function (j) {
     setTimeout(function () {
          console.log(j)
     }, 1000*j);
})(i);

answered Oct 26, 2014 at 8:09

Misha Shblin

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Comments

Sacho · Accepted Answer · 2014-10-26 08:30:34Z

Alongside RGraham's answer, I'd like to point out that when learning javascript, you should try to avoid using function expressions - they make code confusing to read.

Here's what your initial code looks like, without a function expression:

for (var i = 1; i <= 10; i++) {
    setTimeout(doSomething(i), 1000);
}
function doSomething(i) {
    console.log(i);
}

Or an even clearer representation:

for (var i = 1; i <= 10; i++) {
    var k = doSomething(i);
    // doSomething doesn't explicitly return anything, so k === undefined
    setTimeout(k, 1000);
}
function doSomething(i) {
    console.log(i);
}

Also, as I mentioned in my comment, your problem isn't really related to your closure attempt, but rather the way asynchronous operations work in Javascript. The most commonly used ways to sequence asynchronous operations are the continuation passing style/callbacks and promises.

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