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

The behavior you're observing is defined in the ECMA-262 standard

Here's an excerpt from the specification of the << left shift operator:

The production ShiftExpression : ShiftExpression << AdditiveExpression is evaluated as follows:

1. Let lref be the result of evaluating ShiftExpression.
2. Let lval be GetValue(lref).
3. Let rref be the result of evaluating AdditiveExpression.
4. Let rval be GetValue(rref).
5. Let lnum be ToInt32(lval).
6. Let rnum be ToUint32(rval).
7. Let shiftCount be the result of masking out all but the least significant 5 bits of rnum, that is, compute rnum & 0x1F.
8. Return the result of left shifting lnum by shiftCount bits. The result is a signed 32-bit integer.

As you can see, both operands are cast to 32 bit integers. Hence the disappearance of decimal parts.

The same applies for the other bit shift operators. You can find their respective descriptions in section 11.7 Bitwise Shift Operators of the document I linked to.

In this case, the only effect of performing the shift is type conversion. Math.random() returns a floating point value.