I recently found this piece of JavaScript code:
Math.random() * 0x1000000 << 0
I understood that the first part was just generating a random number between 0 and 0x1000000 (== 16777216).
But the second part seemed odd. What's the point of performing a bit-shift by 0? I didn't think that it would do anything. Upon further investigation, however, I noticed that the shift by 0 seemed to truncate the decimal part of the number. Furthermore, it didn't matter if it was a right shift, or a left shift, or even an unsigned right shift.
> 10.12345 << 0
10
> 10.12345 >> 0
10
> 10.12345 >>> 0
10
I tested both with Firefox and Chrome, and the behavior is the same. So, what is the reason for this observation? And is it just a nuance of JavaScript, or does it occur in other languages as well? I thought I understood bit-shifting, but this has me puzzled.
To truncate a number, we miss off digits past a certain point in the number, filling-in zeros if necessary to make the truncated number approximately the same size as the original number. To truncate a number to 1 decimal place, miss off all the digits after the first decimal place.
trunc() The Math. trunc() function returns the integer part of a number by removing any fractional digits.
You're correct; it is used to truncate the value.
The reason >>
works is because it operates only on 32-bit integers, so the value is truncated. (It's also commonly used in cases like these instead of Math.floor
because bitwise operators have a low operator precedence, so you can avoid a mess of parentheses.)
And since it operates only on 32-bit integers, it's also equivalent to a mask with 0xffffffff
after rounding. So:
0x110000000 // 4563402752
0x110000000 >> 0 // 268435456
0x010000000 // 268435456
But that's not part of the intended behaviour since Math.random()
will return a value between 0 and 1.
Also, it does the same thing as | 0
, which is more common.
Math.random()
returns a number between 0 (inclusive) and 1 (exclusive). Multiplying this number with a whole number results in a number that has decimal portion. The <<
operator is a shortcut for eliminating the decimal portion:
The operands of all bitwise operators are converted to signed 32-bit integers in big-endian order and in two's complement format.
The above statements means that the JavaScript engine will implicitly convert both operands of <<
operator to 32-bit integers; for numbers it does so by chopping off the fractional portion (numbers that do not fit 32-bit integer range loose more than just the decimal portion).
And is it just a nuance of JavaScript, or does it occur in other languages as well?
You'll notice similar behavior in loosely typed languages. PHP for example:
var_dump(1234.56789 << 0);
// int(1234)
For strongly types languages, the programs will usually refuse to compile. C# complains like this:
Console.Write(1234.56789 << 0);
// error CS0019: Operator '<<' cannot be applied to operands of type 'double' and 'int'
For these languages, you already have type-casting operators:
Console.Write((int)1234.56789);
// 1234
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