Moreover, I think that some answers are not doing what we should expect from the subject. For example the "Detect Multiple Of Two" detects powers of two, not multiples.
My C++ is a bit rusty, but I think the code is in fact checking if the number is divisible by 2 (i.e. n % 2 == 0).
I think it's using the bitwise and operator (single &) to AND each bit in n and (n-1) and then checking if the least significant bit is 1 or 0. The code would need to shift (<< or >>) to check if the number were a power of 2.
I thought there was another bitwise NOT operator, not the !, but I think, and this is the part I'm hazy on after getting home, that the ! applied to an int is intended to flip each bit. Here's why:
------------
n = 6 = 110
n-1 = 5 = 101
n & (n-1) = 110 & 101 = 100
!(110) = 001 => true
------------
n = 5 = 101
n-1 = 100
n & (n-1) = 101 & 100 = 101
!(101) = 010 => false only if the least significant bit is used for logic decisions... this seems non-portable for some reason, just like using the ! as a bitwise NOT. It may actually be that ! only looks at the least significant bit, but I can't find c++ docs to say one way or another.
Whatever the code actually does, this is exactly the kind of example to use as a poster child for adding just a few comments.
The GP is correct, this is bit-twiddling code to test for a single bit being set in the int, and therefore being a power of 2, not simply a multiple of 2.
Broadly speaking, and ignoring edge cases, if N is a power of 2 then it has a single bit set. Subtracting one unsets that bit, so the AND of N and (N-1) is zero. On the other hand, if N is not a power of two then subtracting 1 leaves the top bit still set, so the AND is non-zero. Therefore taking the boolean NOT gives the right answer.
00010000 Power of 2
00001111 N-1
--------
00000000 -> 0
00010110 Non power of 2
00010101 N-1
--------
00010100 -> Non 0
Thus power of 2 is NOT( N & (N-1) ). Your examples are wrong because "!" is not bit-wise NOT, but boolean NOT. !0 = True, !N = False when N is non-zero.
It's actually wrong on the edge case of N=1, which is a power of 2.
My C++ is a bit rusty, but I think the code is in fact checking if the number is divisible by 2 (i.e. n % 2 == 0).
I think it's using the bitwise and operator (single &) to AND each bit in n and (n-1) and then checking if the least significant bit is 1 or 0. The code would need to shift (<< or >>) to check if the number were a power of 2.
I thought there was another bitwise NOT operator, not the !, but I think, and this is the part I'm hazy on after getting home, that the ! applied to an int is intended to flip each bit. Here's why:
------------
n = 6 = 110
n-1 = 5 = 101
n & (n-1) = 110 & 101 = 100
!(110) = 001 => true
------------
n = 5 = 101
n-1 = 100
n & (n-1) = 101 & 100 = 101
!(101) = 010 => false only if the least significant bit is used for logic decisions... this seems non-portable for some reason, just like using the ! as a bitwise NOT. It may actually be that ! only looks at the least significant bit, but I can't find c++ docs to say one way or another.
Whatever the code actually does, this is exactly the kind of example to use as a poster child for adding just a few comments.