javascript, issue with number rounding

Question

I knew javascript could have rounding issue with divisions, but not with multiplication. How do you solve those?

var p = $('input[name="productsUS"]').val().replace(",", ".");
var t = $('input[name="productsWorld"]').val().replace(",", ".");

if (p >= 0 && t >= 1) {
    var r = p / t;
    r = Math.round(r * 10000) / 10000;
    var aff = (r * 100) + "%";

if p = 100 and t = 57674

r = 0.0017 (ok) and aff = 0.16999999999999998% (arg)

How could I obtain aff = 0.17?

Nop, toFixed() was the solution. I just gave time to people to put up answers and to pick one. — Kraz
– Kraz, Commented Apr 25, 2012 at 16:40

Aidanc · Accepted Answer · 2012-04-25 14:27:14Z

2

("0.16999999999999998").tofixed(2) gives you 0.17.

answered Apr 25, 2012 at 14:27

Aidanc

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gdoron · Accepted Answer · 2012-04-25 14:35:43Z

1

var aff = (r * 100).toFixed(2) + "%";

Live DEMO

toFixed on MDN

edited Apr 25, 2012 at 14:35

answered Apr 25, 2012 at 14:28

gdoron

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Dagg Nabbit · Accepted Answer · 2012-04-25 14:36:54Z

1

If you want to aff to remain a Number instead of being converted to a String, you can use toFixed to work around the precision issues and then "cast" it back to a number using the unary + operator like so:

var n = 0.16999999999999998;

n = +n.toFixed(10); // 0.17

You probably want to use a higher precision than 2 decimal places to avoid rounding issues, I used 10 here.

answered Apr 25, 2012 at 14:36

Dagg Nabbit

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callumacrae · Accepted Answer · 2012-04-25 14:31:41Z

0

That's a bug in JavaScript (although it also affects a few other languages, such as Python), due to the way it stores non-whole numbers being a bit buggy (binary!). The only way to work around it is to round it to two decimal places.

answered Apr 25, 2012 at 14:31

callumacrae

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