Trying some bit manipulation in javascript.

Consider the following:

const n = 4393751543811;
console.log(n.toString(2)) // '111111111100000000000000000000000000000011'
console.log(n & 0b11) // last two bits equal 3
const m = n >> 2; // right shift 2
// The unexpected.
console.log(m.toString(2)) // '0'

The result is 0? The expected output I am looking for after the right shift is:

111111111100000000000000000000000000000011 // pre
001111111111000000000000000000000000000000 // post >> 

How is this accomplished?

CodePudding user response：

Javascript bitwise operators on numbers work "as if" on 32bit integers.

>> (sign-propagating right-shift for numbers) will first convert to a 32-bit integer. If you read linked spec, note specifically

Let int32bit be int modulo 2³².

In other words, all bits above 32 will simply be ignored. For your number, this results in the following:

111111111100000000000000000000000000000011
┗removed━┛┗━━━━━━━━━━━━━━32bit━━━━━━━━━━━━━┛

If you want, you can use BigInt:

const n = 4393751543811n; // note the n-suffix
console.log(n.toString(2))
console.log(n & 0b11n) // for BigInt, all operands must be BigInt
const m = n >> 2n;
// The expected.
console.log(m.toString(2))

The spec for >> on BigInt uses BigInt::leftShift(x, -y), where it in turn states:

Semantics here should be equivalent to a bitwise shift, treating the BigInt as an infinite length string of binary two's complement digits.