Using bitwise OR 0 to floor a number

Question

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.

Answer 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 32-bit 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/or-vs-floor/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.
• 32-bit signed integers only
• Odd Comparative behavior: Math.floor(NaN) === NaN, while (NaN | 0) === 0

Answer 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.

Answer 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

Answer 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.

Answer 5

Javascript represents Number as Double Precision 64-bit 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 0. 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 than 2147483647.

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);
}

Answer 6

• 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

Answer 7

var myNegInt = -1 * Math.pow(2, 32);
var myFloat = 0.010203040506070809;
var my64BitFloat = myNegInt - myFloat;
var trunc1 = my64BitFloat | 0;
var trunc2 = ~~my64BitFloat;
var trunc3 = my64BitFloat ^ 0;
var trunc4 = my64BitFloat - my64BitFloat % 1;
var trunc5 = parseInt(my64BitFloat);
var trunc6 = Math.floor(my64BitFloat);
console.info(my64BitFloat);
console.info(trunc1);
console.info(trunc2);
console.info(trunc3);
console.info(trunc4);
console.info(trunc5);
console.info(trunc6);

IMO: The question "How does it work?", "Does it have any advantages over doing Math.floor?", "Does it have any disadvantages?" pale in comparison to "Is it at all logical to use it for this purpose?"

I think, before you try to get clever with your code, you may want to run these. My advice; just move along, there is nothing to see here. Using bitwise to save a few operations and having that matter to you at all, usually means your code architecture needs work. As far as why it may work sometimes, well a stopped clock is accurate twice a day, that does not make it useful. These operators have their uses, but not in this context.