Right shift operator - Javascript

Question

I'm trying to understand why i.e. Math.random()*255>>0; will skip/remove all the decimals. Same thing happens if I write >>1 or >>2 instead of 0.

I came over another SO-post that said x >> n operator could be looked at as x / 2^n. That still doesn't explain why the decimals goes away.

Any help would be appreciated!

@Phil How is this a duplicate of the link you're referring to? >> and >>> are not the same thing? — Nilzone-
– Nilzone-, Commented Nov 11, 2013 at 0:34
possible duplicate of Why does a shift by 0 truncate the decimal? — user1726343
– user1726343, Commented Nov 11, 2013 at 0:57

Community · Accepted Answer · 2020-06-20 09:12:55Z

4

According to spec, certain numerical operations are required to convert arguments to 32 bit integers first. (http://www.ecma-international.org/ecma-262/5.1/#sec-11.7.2)

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

Let lref be the result of evaluating ShiftExpression.

Let lval be GetValue(lref).

Let rref be the result of evaluating AdditiveExpression.

Let rval be GetValue(rref).

Let lnum be ToInt32(lval). ← The number is converted to a 32 bit integer here

Let rnum be ToUint32(rval).

Let shiftCount be the result of masking out all but the least significant 5 bits of rnum, that is, compute rnum & 0x1F.

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.

edited Jun 20, 2020 at 9:12

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answered Nov 11, 2013 at 0:31

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