Coming from the C programming language, it’s easy to be “negative” about how JavaScript deals with 32-bit integers.

As a newcomer, you quickly learn that JavaScript supports only one numeric data type – 64-bit floats – and you groan.

Then you learn that all the “bitwise” operators (~, |, &, ^, <<, >> and >>>) treat their operands as 32-bit integer values and produce 32-bit integer results, and you breathe a sigh of relief.

But then you start noticing oddities. In C, you can take any 32-bit value, such as -1526726656 (which is equivalent to 0xA5000000), mask it with 0x80808080, and get 0x80000000. However, in JavaScript, you actually get -0x80000000, which, sadly, is not equal to 0x80000000.

To verify, type the following into any JavaScript REPL (eg, Node):

``````> n = -1526726656
-1526726656
> n &= 0x80808080
-2147483648
> n == 0x80000000
false
> n == -0x80000000
true
``````

The sign (bit 31) of every 32-bit result is always extended into the entire 52 “significand” bits of the underlying 64-bit float. And it’s impossible to simply “mask away” those additional sign bits, thanks to a fundamental restriction of JavaScript bitwise operators: they operate only on the low 32 bits.

With one exception: the unsigned right-shift operator. It does more than simply shift zero bits in from the left; it also zeros all the bits above the sign bit. This means that `n >>> 0`, while leaving the low 32 bits unchanged, also clears the upper bits, resulting in a value that is positive, albeit outside the signed 32-bit range. It is equivalent to adding the 33-bit value 0x100000000 to a negative 32-bit number:

``````> n = (n < 0? n + 0x100000000 : n)
2147483648
> n.toString(16)
'80000000'
``````

These operations work because JavaScript is perfectly capable of representing 0x80000000, or any other 32-bit value, as a positive number, but it must use a floating point value to do so. And be careful, because as soon as you perform any bitwise operation on a value with bit 31 set, even an operation as innocuous-looking as:

``````> n |= 0
-2147483648
> n.toString(16)
'-80000000'
``````

the result will be negative again. This is simply how all bitwise operators (except for unsigned right-shift) operate: they truncate the result to a signed 32-bit value.

This might tempt you to think that the right way to write negative 32-bit constants in hex is to simply precede them with a minus sign. But that would be wrong. For example, if you wrote the constant 0x80000080 as “-0x80000080”, JavaScript would treat that as negation of 2147483776, resulting in a value whose low 32 bits are 0x7FFFFF80, not 0x80000080.

The safest way to write a 32-bit constant like 0x80000080 is “0x80000080|0”, which will produce -2147483520. If you write all your negative 32-bit constants that way, then you won’t have to resort to using either unsigned right-shifts or 33-bit addition, which in turn avoids the use of floating point values.

To continue the fun, try setting bit 0 of 0x80000000, which should give you 0x80000001:

``````> n |= 1
-2147483647
> n.toString(16)
'-7fffffff'
``````

WTF? Have all the low 32 bits flipped instead?

Actually, no, this time, I’m pulling your leg. The low 32 bits of the internal value are exactly what you would expect: 0x80000001 (the internal representation is more like 0xFFFFF80000001). But as the MDN Docs explain, for a negative number, toString() returns the positive representation of the number, preceded by a - sign, not the “two’s complement” of the number.

@jeffpar
October 26, 2014 (Updated September 8, 2015)