Why does a shift by 0 truncate the decimal?

Question

I recently found this piece of JavaScript code:

Math.random() * 0x1000000 << 0

I understood that the first part was just generating a

Answer 1

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