What mathematical function does bitwise AND (&) operator do (JS)?

Question

A little context, I was looking at another SO post when I am trying to solve javascript problem to find all possible subsets. I am not asking about the JS challenge, but why it was there and what mathematical significance does it have?

Here is the copy paste of the code from this SO post

var arr = [1, 2, 3];

function generatePowerSet(array) {
  var result = [];
  result.push([]);

  for (var i = 1; i < Math.pow(2, array.length); i++, result.push(subset))
    for (var j = 0, subset = []; j < array.length; j++)
      if (i & Math.pow(2, j))
        subset.push(array[j]);

  return result;
}

console.log(generatePowerSet(arr));

I don't understand what is being accomplished by if (i & Math.pow(2, j)) line. The mozilla doc says that it performs AND comparison on each bits pair. Why is it relevant?

What I mean when I say relevant, for example with LEFT SHIFT, doing a << 1 multiples a by two. Its equivalent mathematical function is multiplying by two if a << b and b is 1. I don't understand what mathematical function & does in this case.

Answer 1

The expression i & Math.pow(2, j) gives a non-zero value when the j^th bit in i is 1 (counting from the least significant bit, which is the 0^th bit).

This can be best explained by example. Let's say i is 10 at a certain moment; in binary: 1010. Now let j be 0. Then:

     i & Math.pow(2, j)
 ==  10 & Math.pow(2, 0)
 ==  10 & 1
 ==  0b1010 & 0b0001
 ==  0b0000

The effect of the second value (0b0001) is filtering: it filters out exactly one bit from the first value. See what happens when j is 1:

     i & Math.pow(2, j)
 ==  10 & Math.pow(2, 1)
 ==  10 & 2
 ==  0b1010 & 0b0010
 ==  0b0010

The if condition will thus be true for this value of j.

As i has two 1 bits, there will be two times that the if condition is true for that particular value of i.