Question

177

setTimeout() - in for loop with random delay

rated 0 times [ 181] [ 4] / answers: 1 / hits: 28452 / 12 Years ago, fri, april 20, 2012, 12:00:00

Possible Duplicate:

Javascript closure inside loops - simple practical example

Seen many posts talking about setTimeout and closures but I'm still not able to pass in a simple for loop counter.

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

  setTimeout(function () {

    console.log(i);

  }, Math.floor(Math.random() * 1000));

}

Gives

5

5

5

5

5

Would like to have

0

1

2

3

4

What's wrong ?

Please don't flame, I thought I have understood the setTimeout() tale but apparently not.

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1 Year ago

answered 12 Years ago pranavrorys · Accepted Answer

You can use a closure to keep a reference to the current value of i within the loop:

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

    (function(i) {

        setTimeout(function () {

            console.log(i);

        }, Math.floor(Math.random() * 1000));

    })(i); //Pass current value into self-executing anonymous function

}

However, this is unlikely to print the numbers in order since you use a random timeout (you could use i * 1000 instead to make the numbers print in ascending order, one second apart).

Here's a working example.