After solving this problem, I looked to some of other's code. In several codes, I found something like this (not exactly this, it's a simplified version of that):

```
int n = 10;
for(int i = 1; i<=n; i+=i& - i;)
{
cout << i << " ";
}
```

**Output:** *1 2 4 8*

I also tried for different values of n and the program outputs powers of 2 less than or equal to n.

But how is it working? What's exactly meant by **i+=i&-i** ?

Thanks for your patience and insight.

i & (-i) is a bit trick for computing the least significant bit of i.

That said, if we are only concerned with powers of 2, i & (-i) is redundant because i & (-i) = i (powers of 2 has only 1 bit set).

If you are wondering why i & (-i) computes the least significant bit, you should learn about how two's complement works.

Wouldn't

`i & 0x1`

do the same? It also isn't susceptible to integer overflow.No that just computes the value of the 0th bit. And how is i & (-i) susceptible to integer overflow?

Oh, I see now that i & (-i) gets the least significant non-zero bit.

As for the overflow, consider the case i == INT_MIN.

Which I guess is irrelevant when i cannot be INT_MIN