Javascript Math.floor function blunder or implementation mystery?

Question

 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?

You know about 0.1 + 0.2 !== 0.3, right? :) Your observation is probably a variation on the theme. — Šime Vidas
– Šime Vidas, Commented Aug 14, 2012 at 18:15
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
– Diego, Commented Aug 14, 2012 at 18:17
See here: jsfiddle.net/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
– Travis J, Commented Aug 14, 2012 at 18:19
@treehau5 i think 15 9s is plenty for whatever you are trying to do — Ibu
– Ibu, Commented Aug 14, 2012 at 18:22

Šime Vidas · Accepted Answer · 2012-08-14 18:26:49Z

The 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.org/wiki/Double-precision_floating-point_format#IEEE_754_double-precision_binary_floating-point_format:_binary64

Jonathan S. · Accepted Answer · 2012-08-14 18:27:11Z

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

Ruan Mendes · Accepted Answer · 2012-08-14 18:21:43Z

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

ajax333221 · Accepted Answer · 2012-08-14 19:19:40Z

See this workaround:

http://jsfiddle.net/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.

Justin Baker · Accepted Answer · 2012-08-16 13:19:12Z

You may pass both an instance of Number, as well as a number literal.

Math.floor(43.99999999999999); Math.floor(new Number(43.99999999999999));

Collectives™ on Stack Overflow

Javascript Math.floor function blunder or implementation mystery?

5 Answers 5

Comments

Comments

Comments

Comments

Comments

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

Comments

Comments

Comments

Comments

Comments

Related