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

According to ECMAScript Language Specification:
http://ecma-international.org/ecma-262/5.1/#sec-11.7.1

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 performing a sign-extending right shift of lnum by shiftCount bits. The most significant bit is propagated. The result is a signed 32-bit integer.