Using bitwise OR 0 to floor a number
A colleague of mine stumbled upon a method to floor float numbers using a bitwise or:
var a = 13.6  0; //a == 13
We were talking about it and wondering a few things.
How does it work? Our theory was that using such an operator casts the number to an integer, thus removing the fractional part
Does it have any advantages over doing Math.floor? Maybe it’s a bit faster? (pun not intended)
Does it have any disadvantages? Maybe it doesn’t work in some cases? Clarity is an obvious one, since we had to figure it out, and well, I’m writting this question.
Thanks.
Solutions/Answers:
Solution 1:
How does it work? Our theory was that using such an operator casts the
number to an integer, thus removing the fractional part
All bitwise operations except unsigned right shift, >>>
, work on signed 32bit integers. So using bitwise operations will convert a float to an integer.
Does it have any advantages over doing Math.floor? Maybe it’s a bit
faster? (pun not intended)
http://jsperf.com/orvsfloor/2 seems slightly faster
Does it have any disadvantages? Maybe it doesn’t work in some cases?
Clarity is an obvious one, since we had to figure it out, and well,
I’m writting this question.
 Will not pass jsLint.
 32bit signed integers only
 Odd Comparative behavior:
Math.floor(NaN) === NaN
, while(NaN  0) === 0
Solution 2:
This is truncation as opposed to flooring. Howard’s answer is sort of correct; But I would add that Math.floor
does exactly what it is supposed to with respect to negative numbers. Mathematically, that is what a floor is.
In the case you described above, the programmer was more interested in truncation or chopping the decimal completely off. Although, the syntax they used sort of obscures the fact that they are converting the float to an int.
Solution 3:
In ECMAScript 6, the equivalent of 0
is Math.trunc, kind of I should say:
Returns the integral part of a number by removing any fractional digits. It just truncate the dot and the digits behind it, no matter whether the argument is a positive number or a negative number.
Math.trunc(13.37) // 13
Math.trunc(42.84) // 42
Math.trunc(0.123) // 0
Math.trunc(0.123) // 0
Math.trunc("1.123")// 1
Math.trunc(NaN) // NaN
Math.trunc("foo") // NaN
Math.trunc() // NaN
Solution 4:
Your first point is correct. The number is cast to an integer and thus any decimal digits are removed. Please note, that Math.floor
rounds to the next integer towards minus infinity and thus gives a different result when applied to negative numbers.
Solution 5:

The specs say that it is converted to an integer:
Let lnum be ToInt32(lval).

Performance: this has been tested at jsperf before.
note: dead link to spec removed
Solution 6:
Javascript represents Number
as Double Precision 64bit Floating numbers.
Math.floor
works with this in mind.
Bitwise operations work in 32bit signed integers. 32bit signed integers use first bit as negative signifier and the other 31 bits are the number. Because of this, the min and max number allowed 32bit signed numbers are 2,147,483,648 and 2147483647 (0x7FFFFFFFF), respectively.
So when you’re doing  0
, you’re essentially doing is & 0xFFFFFFFF
. This means, any number that is represented as 0x80000000 (2147483648) or greater will return as a negative number.
For example:
// Safe
(2147483647.5918 & 0xFFFFFFFF) === 2147483647
(2147483647 & 0xFFFFFFFF) === 2147483647
(200.59082098 & 0xFFFFFFFF) === 200
(0X7FFFFFFF & 0xFFFFFFFF) === 0X7FFFFFFF
// Unsafe
(2147483648 & 0xFFFFFFFF) === 2147483648
(2147483649 & 0xFFFFFFFF) === 2147483647
(0x80000000 & 0xFFFFFFFF) === 2147483648
(3000000000.5 & 0xFFFFFFFF) === 1294967296
Also. Bitwise operations don’t “floor”. They truncate, which is the same as saying, they round closest to . Once you go around to negative numbers,
Math.floor
rounds down while bitwise start rounding up.
As I said before, Math.floor
is safer because it operates with 64bit floating numbers. Bitwise is faster, yes, but limited to 32bit signed scope.
To summarize:
 Bitwise works the same if you work from
0 to 2147483647
.  Bitwise is 1 number off if you work from
2147483647 to 0
.  Bitwise is completely different for numbers less than
2147483648
and greater than2147483647
.
If you really want to tweak performance and use both:
function floor(n) {
if (n >= 0 && n < 0x80000000) {
return n & 0xFFFFFFFF;
}
if (n > 0x80000000 && n < 0) {
return (n  1) & 0xFFFFFFFF;
}
return Math.floor(n);
}
Just to add Math.trunc
works like bitwise operations. So you can do this:
function trunc(n) {
if (n > 0x80000000 && n < 0x80000000) {
return n & 0xFFFFFFFF;
}
return Math.trunc(n);
}