Dealing with float precision in Javascript [duplicate]

Answer 1

Check out this link.. It helped me a lot.

http://www.w3schools.com/jsref/jsref_toprecision.asp

The toPrecision(no_of_digits_required) function returns a string so don't forget to use the parseFloat() function to convert to decimal point of required precision.

Answer 2

From this post: How to deal with floating point number precision in JavaScript?

You have a few options:

• Use a special datatype for decimals, like decimal.js
• Format your result to some fixed number of significant digits, like this: (Math.floor(y/x) * x).toFixed(2)
• Convert all your numbers to integers

Answer 3

You could do something like this:

> +(Math.floor(y/x)*x).toFixed(15);
1.2

Answer 4

Tackling this task, I'd first find the number of decimal places in x, then round y accordingly. I'd use:

y.toFixed(x.toString().split(".")[1].length);

It should convert x to a string, split it over the decimal point, find the length of the right part, and then y.toFixed(length) should round y based on that length.

Answer 5

> var x = 0.1
> var y = 0.2
> var cf = 10
> x * y
0.020000000000000004
> (x * cf) * (y * cf) / (cf * cf)
0.02

Quick solution:

var _cf = (function() {
  function _shift(x) {
    var parts = x.toString().split('.');
    return (parts.length < 2) ? 1 : Math.pow(10, parts[1].length);
  }
  return function() { 
    return Array.prototype.reduce.call(arguments, function (prev, next) { return prev === undefined || next === undefined ? undefined : Math.max(prev, _shift (next)); }, -Infinity);
  };
})();

Math.a = function () {
  var f = _cf.apply(null, arguments); if(f === undefined) return undefined;
  function cb(x, y, i, o) { return x + f * y; }
  return Array.prototype.reduce.call(arguments, cb, 0) / f;
};

Math.s = function (l,r) { var f = _cf(l,r); return (l * f - r * f) / f; };

Math.m = function () {
  var f = _cf.apply(null, arguments);
  function cb(x, y, i, o) { return (x*f) * (y*f) / (f * f); }
  return Array.prototype.reduce.call(arguments, cb, 1);
};

Math.d = function (l,r) { var f = _cf(l,r); return (l * f) / (r * f); };

> Math.m(0.1, 0.2)
0.02

You can check the full explanation here.