According to Wikipedia, the modulo operator (remainder of integer division) on n should yield a result between 0 and n-1.
This is indeed the case in python:
print(-1%5) # outputs 4
In Ruby:
puts -1%5 # outputs 4
In Haskell
main = putStrLn $ show $ mod (-1) 5
But inJavascript:
console.log(-1%5);
The result is -1 !!
Why is that ? is there logic behind this design decision ?
While -1 is equivalent to 4 modulo 5, negative values makes it harder to use as indices.
For example:
arr[x%5]
would always be a valid expression in python and Ruby when the array length is more than 5, but in the Javascript, this is an exception waiting to happen.
If by "modulus" you understand the remainder of the Euclidean division as common in mathematics, the % operator is not the modulo operator. It rather is called the remainder operator, whose result always has the same sign as the dividend. (In Haskell, the rem function does this).
This behaviour is pretty common amongst programming languages, notably also in C and Java from which the arithmetic conventions of JavaScript were inspired.
Note that while in most languages, ‘%’ is a remainder operator, in some (e.g. Python, Perl) it is a modulo operator. For two values of the same sign, the two are equivalent, but when the dividend and divisor are of different signs, they give different results. To obtain a modulo in JavaScript, in place of a % n, use ((a % n ) + n ) % n
see
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Remainder
Related
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.
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
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.
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
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) {
const bitFloored = n & 0xFFFFFFFF;
if (bitFloored === n) return n;
return bitFloored - 1;
}
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);
}
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.
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
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.
I was looking at code from Mozilla that add a filter method to Array and it had a line of code that confused me.
var len = this.length >>> 0;
I have never seen >>> used in JavaScript before. What is it and what does it do?
It doesn't just convert non-Numbers to Number, it converts them to Numbers that can be expressed as 32-bit unsigned ints.
Although JavaScript's Numbers are double-precision floats(*), the bitwise operators (<<, >>, &, | and ~) are defined in terms of operations on 32-bit integers. Doing a bitwise operation converts the number to a 32-bit signed int, losing any fractions and higher-place bits than 32, before doing the calculation and then converting back to Number.
So doing a bitwise operation with no actual effect, like a rightward-shift of 0 bits >>0, is a quick way to round a number and ensure it is in the 32-bit int range. Additionally, the triple >>> operator, after doing its unsigned operation, converts the results of its calculation to Number as an unsigned integer rather than the signed integer the others do, so it can be used to convert negatives to the 32-bit-two's-complement version as a large Number. Using >>>0 ensures you've got an integer between 0 and 0xFFFFFFFF.
In this case this is useful because ECMAScript defines Array indexes in terms of 32 bit unsigned ints. So if you're trying to implement array.filter in a way that exactly duplicates what the ECMAScript Fifth Edition standard says, you would cast the number to 32-bit unsigned int like this.
In reality there's little practical need for this as hopefully people aren't going to be setting array.length to 0.5, -1, 1e21 or 'LEMONS'.
Summary:
1>>>0 === 1
-1>>>0 === 0xFFFFFFFF -1>>0 === -1
1.7>>>0 === 1
0x100000002>>>0 === 2
1e21>>>0 === 0xDEA00000 1e21>>0 === -0x21600000
Infinity>>>0 === 0
NaN>>>0 === 0
null>>>0 === 0
'1'>>>0 === 1
'x'>>>0 === 0
Object>>>0 === 0
(*: well, they're defined as behaving like floats. It wouldn't surprise me if some JavaScript engine actually used ints when it could, for performance reasons. But that would be an implementation detail you wouldn't get to take any advantage of.)
The unsigned right shift operator is used in the all the array extra's method implementations of Mozilla, to ensure that the length property is a unsigned 32-bit integer.
The length property of array objects is described in the specification as:
Every Array object has a length property whose value is always a nonnegative integer less than 232.
This operator is the shortest way to achieve it, internally array methods use the ToUint32 operation, but that method is not accessible and exist on the specification for implementation purposes.
The Mozilla array extras implementations try to be ECMAScript 5 compliant, look at the description of the Array.prototype.indexOf method (§ 15.4.4.14):
1. Let O be the result of calling ToObject passing the this value
as the argument.
2. Let lenValue be the result of calling the [[Get]] internal method of O with
the argument "length".
3. Let len be ToUint32(lenValue).
....
As you can see, they just want to reproduce the behavior of the ToUint32 method to comply with the ES5 spec on an ES3 implementation, and as I said before, the unsigned right shift operator is the easiest way.
That is the unsigned right bit shift operator. The difference between this and the signed right bit shift operator, is that the unsigned right bit shift operator (>>>) fills with zeroes from the left, and the signed right bit shift operator (>>) fills with the sign bit, thus preserving the sign of the numerical value when shifted.
Driis has sufficiently explained what the operator is and what it does. Here's the meaning behind it/why it was used:
Shifting any direction by 0 does returns the original number and will cast null to 0. It seems that the example code that you are looking at is using this.length >>> 0 to ensure that len is numeric even if this.length is not defined.
For many people, bitwise operations are unclear (and Douglas Crockford/jslint suggests against using such things). It doesn't mean that its wrong to do, but more favorable and familiar methods exist to make code more readable. A more clear way to ensure that len is 0 is either of the following two methods.
// Cast this.length to a number
var len = +this.length;
or
// Cast this.length to a number, or use 0 if this.length is
// NaN/undefined (evaluates to false)
var len = +this.length || 0;
>>> is the unsigned right shift operator (see p. 76 of the JavaScript 1.5 specification), as opposed to the >>, the signed right shift operator.
>>> changes the results of shifting negative numbers because it does not preserve the sign bit when shifting. The consequences of this is can be understood by example, from an interpretter:
$ 1 >> 0
1
$ 0 >> 0
0
$ -1 >> 0
-1
$ 1 >>> 0
1
$ 0 >>> 0
0
$ -1 >>> 0
4294967295
$(-1 >>> 0).toString(16)
"ffffffff"
$ "cabbage" >>> 0
0
So what is probably intended to be done here is to get the length, or 0 if the length is undefined or not an integer, as per the "cabbage" example above. I think in this case it is safe to assume that this.length will never be < 0. Nevertheless, I would argue that this example is a nasty hack, for two reasons:
The behavior of <<< when using negative numbers, a side-effect probably not intended (or likely to occur) in the example above.
The intention of the code is not obvious, as the existence of this question verifies.
Best practice is probably to use something more readable unless performance is absolutely critical:
isNaN(parseInt(foo)) ? 0 : parseInt(foo)
Two reasons:
The result of >>> is an "integral"
undefined >>> 0 = 0 (since JS will try and coerce the LFS to numeric context, this will work for "foo" >>> 0, etc. as well)
Remember that numbers in JS have an internal-representation of double.
It's just a "quick" way of basic input sanity for length.
However, -1 >>> 0 (oops, likely not a desired length!)
Sample Java Code below explains well:
int x = 64;
System.out.println("x >>> 3 = " + (x >>> 3));
System.out.println("x >> 3 = " + (x >> 3));
System.out.println(Integer.toBinaryString(x >>> 3));
System.out.println(Integer.toBinaryString(x >> 3));
Output is the following:
x >>> 3 = 536870904
x >> 3 = -8
11111111111111111111111111000
11111111111111111111111111111000
I am trying to understand Javascript logical operators and came across 2 statements with seeminlgy similar functionality and trying to understand the difference. So, What's the difference between these 2 lines of code in Javascript?
For a number x,
x >>>= 0;
x &= 0x7fffffff;
If I understand it correctly, they both should give unsigned 32 bit output. However, for same negative value of x (i.e. most significant bit always 1 in both case), I get different outputs, what am I missing?
Thanks
To truncate a number to 32 bits, the simplest and most common method is to use the "|" bit-wise operator:
x |= 0;
JavaScript always considers the result of any 32-bit computation to be negative if the highest bit (bit 31) is set. Don't let that bother you. And don't clear bit 31 in an attempt to make it positive; that incorrectly alters the value.
To convert a negative 32-bit number as a positive value (a value in the range 0 to 4294967295), you can do this:
x = x < 0? x + 0x100000000 : x;
By adding a 33-bit value, automatic sign-extension of bit 31 is inhibited. However, the result is now outside the signed 32-bit range.
Another (tidier) solution is to use the unsigned right-shift operator with a zero shift count:
x >>>= 0;
Technically, all JavaScript numbers are 64-bit floating-point values, but in reality, as long as you keep numbers within the signed 32-bit range, you make it possible for JavaScript runtimes to optimize your code using 32-bit integer operations.
Be aware that when you convert a negative 32-bit value to a positive value using either of above methods, you have essentially produced a 33-bit value, which may defeat any 32-bit optimizations your JavaScript engine uses.
I was looking at code from Mozilla that add a filter method to Array and it had a line of code that confused me.
var len = this.length >>> 0;
I have never seen >>> used in JavaScript before. What is it and what does it do?
It doesn't just convert non-Numbers to Number, it converts them to Numbers that can be expressed as 32-bit unsigned ints.
Although JavaScript's Numbers are double-precision floats(*), the bitwise operators (<<, >>, &, | and ~) are defined in terms of operations on 32-bit integers. Doing a bitwise operation converts the number to a 32-bit signed int, losing any fractions and higher-place bits than 32, before doing the calculation and then converting back to Number.
So doing a bitwise operation with no actual effect, like a rightward-shift of 0 bits >>0, is a quick way to round a number and ensure it is in the 32-bit int range. Additionally, the triple >>> operator, after doing its unsigned operation, converts the results of its calculation to Number as an unsigned integer rather than the signed integer the others do, so it can be used to convert negatives to the 32-bit-two's-complement version as a large Number. Using >>>0 ensures you've got an integer between 0 and 0xFFFFFFFF.
In this case this is useful because ECMAScript defines Array indexes in terms of 32 bit unsigned ints. So if you're trying to implement array.filter in a way that exactly duplicates what the ECMAScript Fifth Edition standard says, you would cast the number to 32-bit unsigned int like this.
In reality there's little practical need for this as hopefully people aren't going to be setting array.length to 0.5, -1, 1e21 or 'LEMONS'.
Summary:
1>>>0 === 1
-1>>>0 === 0xFFFFFFFF -1>>0 === -1
1.7>>>0 === 1
0x100000002>>>0 === 2
1e21>>>0 === 0xDEA00000 1e21>>0 === -0x21600000
Infinity>>>0 === 0
NaN>>>0 === 0
null>>>0 === 0
'1'>>>0 === 1
'x'>>>0 === 0
Object>>>0 === 0
(*: well, they're defined as behaving like floats. It wouldn't surprise me if some JavaScript engine actually used ints when it could, for performance reasons. But that would be an implementation detail you wouldn't get to take any advantage of.)
The unsigned right shift operator is used in the all the array extra's method implementations of Mozilla, to ensure that the length property is a unsigned 32-bit integer.
The length property of array objects is described in the specification as:
Every Array object has a length property whose value is always a nonnegative integer less than 232.
This operator is the shortest way to achieve it, internally array methods use the ToUint32 operation, but that method is not accessible and exist on the specification for implementation purposes.
The Mozilla array extras implementations try to be ECMAScript 5 compliant, look at the description of the Array.prototype.indexOf method (§ 15.4.4.14):
1. Let O be the result of calling ToObject passing the this value
as the argument.
2. Let lenValue be the result of calling the [[Get]] internal method of O with
the argument "length".
3. Let len be ToUint32(lenValue).
....
As you can see, they just want to reproduce the behavior of the ToUint32 method to comply with the ES5 spec on an ES3 implementation, and as I said before, the unsigned right shift operator is the easiest way.
That is the unsigned right bit shift operator. The difference between this and the signed right bit shift operator, is that the unsigned right bit shift operator (>>>) fills with zeroes from the left, and the signed right bit shift operator (>>) fills with the sign bit, thus preserving the sign of the numerical value when shifted.
Driis has sufficiently explained what the operator is and what it does. Here's the meaning behind it/why it was used:
Shifting any direction by 0 does returns the original number and will cast null to 0. It seems that the example code that you are looking at is using this.length >>> 0 to ensure that len is numeric even if this.length is not defined.
For many people, bitwise operations are unclear (and Douglas Crockford/jslint suggests against using such things). It doesn't mean that its wrong to do, but more favorable and familiar methods exist to make code more readable. A more clear way to ensure that len is 0 is either of the following two methods.
// Cast this.length to a number
var len = +this.length;
or
// Cast this.length to a number, or use 0 if this.length is
// NaN/undefined (evaluates to false)
var len = +this.length || 0;
>>> is the unsigned right shift operator (see p. 76 of the JavaScript 1.5 specification), as opposed to the >>, the signed right shift operator.
>>> changes the results of shifting negative numbers because it does not preserve the sign bit when shifting. The consequences of this is can be understood by example, from an interpretter:
$ 1 >> 0
1
$ 0 >> 0
0
$ -1 >> 0
-1
$ 1 >>> 0
1
$ 0 >>> 0
0
$ -1 >>> 0
4294967295
$(-1 >>> 0).toString(16)
"ffffffff"
$ "cabbage" >>> 0
0
So what is probably intended to be done here is to get the length, or 0 if the length is undefined or not an integer, as per the "cabbage" example above. I think in this case it is safe to assume that this.length will never be < 0. Nevertheless, I would argue that this example is a nasty hack, for two reasons:
The behavior of <<< when using negative numbers, a side-effect probably not intended (or likely to occur) in the example above.
The intention of the code is not obvious, as the existence of this question verifies.
Best practice is probably to use something more readable unless performance is absolutely critical:
isNaN(parseInt(foo)) ? 0 : parseInt(foo)
Two reasons:
The result of >>> is an "integral"
undefined >>> 0 = 0 (since JS will try and coerce the LFS to numeric context, this will work for "foo" >>> 0, etc. as well)
Remember that numbers in JS have an internal-representation of double.
It's just a "quick" way of basic input sanity for length.
However, -1 >>> 0 (oops, likely not a desired length!)
Sample Java Code below explains well:
int x = 64;
System.out.println("x >>> 3 = " + (x >>> 3));
System.out.println("x >> 3 = " + (x >> 3));
System.out.println(Integer.toBinaryString(x >>> 3));
System.out.println(Integer.toBinaryString(x >> 3));
Output is the following:
x >>> 3 = 536870904
x >> 3 = -8
11111111111111111111111111000
11111111111111111111111111111000
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.
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
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.
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
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) {
const bitFloored = n & 0xFFFFFFFF;
if (bitFloored === n) return n;
return bitFloored - 1;
}
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);
}
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.
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
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.