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Finding if a number is a power of 2

Just out of curiosity, how can you tell if a number x is a power of two (x = 2^n) without using recursion.

Thanks

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Answer

One way is to use bitwise AND. If a number $x is a power of two (e.g., 8=1000), it will have no bits in common with its predecessor (7=0111). So you can write:

($x & ($x - 1)) == 0

Note: This will give a false positive for $x == 0.

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