this post was submitted on 21 Dec 2024
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Because the abs(3) == 3 is true and that isn't even.
An even number of flips would be true and an odd number of flips would be false which works out.
I was thinking a bitwise & or converting it to a string and testing if the right most character is 0, 2, 4, 6, 8 would be panic mode solutions too.
you might be able to do it with a bitwise op? My track record tonight is not great so I'm not going to comment. Have a look at @ImplyingImplactions comment for a loopless solution
Bitwise and with 0x1. If result is 0, it's even. Least significant bit is always 1 for odd numbers.