shanto_bangladesh's blog

Powers of 2 using & . But how?
By shanto_bangladesh, history, 3 years ago, In English

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.

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3 years ago, # |
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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.

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    3 years ago, # ^ |
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    Wouldn't i & 0x1 do the same? It also isn't susceptible to integer overflow.

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