2

Please see code snippet, javascript return unexpected value when calculating numbers. I'm using newest firefox. Is there a way I could always get correct value ?

console.log(100+59.59, "expected 159.59"); 
console.log(200+59.59, "expected 259.59");
console.log(300+59.59, "expected 359.59");
console.log(400+59.59, "expected 459.59");
console.log(500+59.59, "expected 559.59");
console.log(600+286.59, "expected 886.59");
console.log(700+286.59, "expected 986.59");
RoliCo
  • 108
  • 1
  • 7

2 Answers2

0
function FloatAdd(arg1, arg2)
{
  var r1, r2, m;
  try { r1 = arg1.toString().split(".")[1].length; } catch (e) { r1 = 0; }
  try { r2 = arg2.toString().split(".")[1].length; } catch (e) { r2 = 0; }
  m = Math.pow(10, Math.max(r1, r2));
  return (FloatMul(arg1, m) + FloatMul(arg2, m)) / m;
}

function FloatSubtraction(arg1, arg2)
{
  var r1, r2, m, n;
  try { r1 = arg1.toString().split(".")[1].length } catch (e) { r1 = 0 }
  try { r2 = arg2.toString().split(".")[1].length } catch (e) { r2 = 0 }
  m = Math.pow(10, Math.max(r1, r2));
  n = (r1 >= r2) ? r1 : r2;
  return ((arg1 * m - arg2 * m) / m).toFixed(n);
}

function FloatMul(arg1, arg2)
{
  var m = 0, s1 = arg1.toString(), s2 = arg2.toString();
  try { m += s1.split(".")[1].length; } catch (e) { }
  try { m += s2.split(".")[1].length; } catch (e) { }
  return Number(s1.replace(".", "")) * Number(s2.replace(".", "")) / Math.pow(10, m);
}

function FloatDiv(arg1, arg2)
{
  var t1 = 0, t2 = 0, r1, r2;
  try { t1 = arg1.toString().split(".")[1].length } catch (e) { }
  try { t2 = arg2.toString().split(".")[1].length } catch (e) { }
  with (Math)
  {
    r1 = Number(arg1.toString().replace(".", ""))
    r2 = Number(arg2.toString().replace(".", ""))
    return (r1 / r2) * pow(10, t2 - t1);
  }
}

reference http://kevintsengtw.blogspot.tw/2011/09/javascript.html

This question already has an answer here:
Is floating point math broken?

So the quick way is to convert to string then change to float again to fix it

Community
  • 1
  • 1
Joseph M Tsai
  • 568
  • 3
  • 12
0

If you're sure that you don't need more than x decimal places, you can use the toFixed() method, like so:

console.log((100+59.59).toFixed(2), "expected 159.59"); 
console.log((200+59.59).toFixed(2), "expected 259.59");
console.log((300+59.59).toFixed(2), "expected 359.59");
console.log((400+59.59).toFixed(2), "expected 459.59");
console.log((500+59.59).toFixed(2), "expected 559.59");
console.log((600+286.59).toFixed(2), "expected 886.59");
console.log((700+286.59).toFixed(2), "expected 986.59");

The problem here is not Javascript, but the limitations of floating point numbers in binary. 0.59 cannot be perfectly represented with a 64 bit floating point number (it actually works out to 0.58999999999999996891375531049561686813831329345703125 when converting back to decimal).

Kyle G.
  • 870
  • 2
  • 10
  • 22