document.writeln(Math.floor(43.9));
produces 43 in the browser.
document.writeln(Math.floor(43.9999));
produces 43
document.writeln(Math.floor(43.999999999999));
again 43
However,
document.writeln(Math.floor(43.99999999999999));
produces 44.
The magic number of 9's after the decimal point seems to be 15*.
Why is this?
Furthermore, Does the Math.floor function accept the number as a number object, or a number value?
document.writeln(Math.floor(43.9));
produces 43 in the browser.
document.writeln(Math.floor(43.9999));
produces 43
document.writeln(Math.floor(43.999999999999));
again 43
However,
document.writeln(Math.floor(43.99999999999999));
produces 44.
The magic number of 9's after the decimal point seems to be 15*.
Why is this?
Furthermore, Does the Math.floor function accept the number as a number object, or a number value?
Share Improve this question edited Aug 15, 2012 at 20:34 darethas asked Aug 14, 2012 at 18:13 darethasdarethas 7,6673 gold badges25 silver badges32 bronze badges 12-
3
You know about
0.1 + 0.2 !== 0.3, right?:)Your observation is probably a variation on the theme. – Šime Vidas Commented Aug 14, 2012 at 18:15 - 2 I assume (not sure) that the function expects a float and in the first number which returns 44 (the one with 14 9's) exceeds the float size, so your constant number is being rounded and then passed to the function. Again, just a guess! – Diego Commented Aug 14, 2012 at 18:17
- 1 Btw, the magic number is 15 in my Firebug console – Šime Vidas Commented Aug 14, 2012 at 18:17
- 3 See here: jsfiddle/k7huV/2 , you can see that 43.99 (15 9s) is actually stored as 44. So this is not a rounding issue in the math algorithm but instead in the way the number is stored. – Travis J Commented Aug 14, 2012 at 18:19
- 1 @treehau5 i think 15 9s is plenty for whatever you are trying to do – Ibu Commented Aug 14, 2012 at 18:22
5 Answers
Reset to default 7The IEEE 754 double-precision binary floating-point format (which is what JavaScript uses for its Number type) gives you a precision of 15 - 17 significant decimal digits.
This gives from 15 - 17 significant decimal digits precision. If a decimal string with at most 15 significant decimal is converted to IEEE 754 double precision and then converted back to the same number of significant decimal, then the final string should match the original; and if an IEEE 754 double precision is converted to a decimal string with at least 17 significant decimal and then converted back to double, then the final number must match the original [1].
Source: http://en.wikipedia/wiki/Double-precision_floating-point_format#IEEE_754_double-precision_binary_floating-point_format:_binary64
Double-precision floating point numbers (a.k.a. doubles) are capable of storing a very wide range of values but only with limited precision—15–17 significant digits. If you do the following, you'll see what happens:
var x = 43.99999999999999;
var y = 43.999999999999999;
document.writeln(x); // 43.99999999999999
document.writeln(y); // 44
document.writeln(Math.floor(x)); // 43
document.writeln(Math.floor(y)); // 44
You'll see the same in other languages as well. For example, PHP:
echo floor(43.99999999999999); // 43
echo floor(43.999999999999999); // 44
In Chrome, if I just type 43.99999999999999999999999 into the console, it outputs 44, that's probably what you're running into. Floating points are approximations
See this workaround:
http://jsfiddle/k7huV/3/
var x = 43;
for( var i = 0; i < 16; i++ ){
if( i == 0 ) x += ".";
x += "9";
}
document.writeln(x);
document.writeln(parseInt(x));
Output: 43.9999999999999999 43
Correctly floors 43.999999999999999 to 43.
You may pass both an instance of Number, as well as a number literal.
Math.floor(43.99999999999999);
Math.floor(new Number(43.99999999999999));