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